The conics again, this time tilted. Derives the rotation equations from polar coordinates and the sum formulas, uses them to convert both points and equations into a rotated frame, and finds the angle that eliminates the cross term — the one whose cotangent of twice it equals A minus C over B. Closes with the discriminant, which classifies a general second-degree equation as hyperbola, parabola or ellipse without performing any rotation at all.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 11 — Applications of Trigonometry
§11.6 Hooked on Conics Again, pp. 973-981
Objectives
Five outcomes, and the last one is a shortcut worth having before the other four.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 973-981 — the pages these objectives are drawn from
Warm-up
The equation xy equal to 2 has a familiar graph, but it does not look like any of the standard conic forms.
Discussion prompt
Sketch xy equal to 2 and say what shape it is. Then try to match it against the standard equation of a hyperbola.
Hint: What does the graph look like in the first and third quadrants?
Answer:
It is a hyperbola, with the two coordinate axes as its asymptotes and branches in the first and third quadrants. Nothing about the picture is unusual.
But no standard form matches it, because every standard form has its axes along the coordinate axes, and this one's are along the diagonals. The equation contains an xy term and no squared terms at all.
It is in fact the hyperbola x squared minus y squared equal to 4, rotated counter-clockwise by 45 degrees. This lesson makes that statement precise and gives the machinery to undo any such rotation.
Concept
Rotate the axes about the origin. A point keeps its distance from the origin and only its angle is re-measured, differing by exactly the rotation. Expanding with the sum formulas converts between the two sets of coordinates.
Theorem 11.9 — If the axes are rotated counter-clockwise through an angle theta, then x equals x-prime cosine theta minus y-prime sine theta and y equals x-prime sine theta plus y-prime cosine theta; and conversely x-prime equals x cosine theta plus y sine theta with y-prime equal to negative x sine theta plus y cosine theta.
\[ \begin{aligned} x &= x'\cos(\theta) - y'\sin(\theta) \\ y &= x'\sin(\theta) + y'\cos(\theta) \end{aligned} \]
The first system converts equations from the old variables into the new ones; the second converts points. Which you need depends on whether you are moving a curve or a location.
Figure (svg): The original axes with a second pair rotated counter-clockwise through an angle, and a single point carrying coordinates in both systems
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 973-974
Section
Section 1
Concept
Write the point in polar form in each frame. Both use the same distance, since the origin is shared, and the angles differ by the rotation. Expanding with the sum formulas gives the theorem.
The reverse system comes from solving the first for x-prime and y-prime, which the book does with a two by two matrix whose determinant is the Pythagorean Identity and therefore 1.
Figure (svg): The original axes with a second pair rotated counter-clockwise through an angle, and a single point carrying coordinates in both systems
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 973-974
Picture it
One converts equations and one converts points.
Figure (svg): The two systems of rotation equations, one converting from the new coordinates to the old and one from the old to the new
Notice the second system is the first with theta replaced by its negative, which is exactly what undoing a rotation should look like.
Worked example
The whole proof, using nothing but the sum formula for cosine.
\[ \text{Derive } x = x'\cos(\theta) - y'\sin(\theta). \]
Write x in the old frame
Why: The angle from the old x-axis is theta plus phi.
\[ x = r \cos(\theta + p h) \]
Expand with the sum formula
Why: Cosine of a sum.
\[ = r \cos \theta \cos p h - r \sin \theta \sin p h \]
Regroup the factors
Why: Pair each r with the phi function.
\[ = (r \cos p h) \cos \theta - (r \sin p h) \sin \theta \]
Recognise the brackets
Why: They are the new coordinates.
\[ x = x' \cos \theta - y' \sin \theta \]
Figure (svg): The solution to Worked example the derivation shown as a ladder of expressions, one row per legal move
\[ x = x'\cos(\theta) - y'\sin(\theta) \]
Verify: test with no rotation
Why: Setting theta to zero gives x equal to x-prime times 1 minus y-prime times 0, so x equals x-prime. With no rotation the two frames coincide, which is what the formula reports.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 973-974
Sorting
One converts equations and one converts points.
Sort into buckets
Sort each task.
Worked example
Example 11.6.1, part 1. The axes turn through pi over 3.
\[ \text{With } \theta = \tfrac{\pi}{3}, \text{ find the new coordinates of } P(2, -4). \]
Use the old-to-new system
Why: The second pair of equations.
\[ x' = x \cos \theta + y \sin \theta \]
Substitute for x-prime
Why: Two cosine 60 plus negative four sine 60.
\[ x' = 1 - 2 \sqrt{3} \]
Substitute for y-prime
Why: Negative two sine 60 plus negative four cosine 60.
\[ y' = -\sqrt{3} - 2 \]
Approximate for the sketch
Why: To check the answer looks right.
\[ (-2.46, -3.73) \]
Figure (svg): The solution to Worked example rotating a point shown as a ladder of expressions, one row per legal move
\[ P(x', y') = \left(1 - 2\sqrt{3}, \; -2 - \sqrt{3}\right) \approx (-2.46, -3.73) \]
Verify: convert back
Why: Using the new-to-old system, x equals one minus two root three times one half, minus negative two minus root three times root three over two, which works out to 2. The same computation for y gives negative 4. The round trip returns the original point.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 975-975
Trap
\[ \text{convert } x^2 - y^2 = 4 \text{ by substituting } x' = x\cos\theta + y\sin\theta \]
Substitute the old-to-new formulas into the equation
Why: They are the ones expressing the new variables, which is where the equation is going.
But substituting into an equation replaces the variables that appear in it. The equation contains x and y, so it is x and y that must be replaced — by the new-to-old formulas.
\[ x = x'\cos\theta - y'\sin\theta, \quad y = x'\sin\theta + y'\cos\theta \]
Substitute expressions for the variables that actually appear
Why: The equation is in x and y, so replace x and y.
The rule that settles it every time: to convert an equation, substitute for the variables it contains; to convert a point, use the formulas that produce the coordinates you want. Those are opposite systems, which is why the theorem states both.
Faded example
The axes turn through pi over 4. Find the new coordinates of (3, 5).
Fill in the blanks
x' = 3\cos\tfrac82 + 5\sin\tfrac______ = ___\tfrac___}___, \quad y' = -3\sin\tfrac______ + 5\cos\tfrac______ = ___\tfrac___}___
Why: Both the sine and cosine of pi over 4 are root two over two, so the coefficients simply add and subtract: 3 plus 5 is 8, and negative 3 plus 5 is 2. A 45 degree rotation is the one case where both trigonometric values coincide, which is why it produces such clean arithmetic.
Prediction
A point is rotated to a new frame and then the frame is rotated back.
Predict first
What must the composition of the two systems be?
Correct: The identity.
Why: Rotating by theta and then by negative theta returns every point to where it started, so the composition must leave every coordinate unchanged. This is the algebraic content of the determinant being 1 and the inverse matrix being the transpose: the second system is the first with theta negated, and the two undo each other exactly. It is also the check to run on any rotation calculation.
Socratic
Deriving the equations directly from the geometry would take a page of similar triangles.
Discussion prompt
What does the polar description supply that the rectangular one does not?
Hint: What stays the same when the axes rotate?
Answer:
Polar coordinates separate what changes from what does not. Rotating the axes leaves the distance from the origin completely untouched and changes only the angle, and it changes that angle by a fixed amount.
So in polar form the transformation is subtract theta from the angle and leave r alone — as simple as a transformation can be. All the complexity of the rectangular formulas is the cost of translating that one simple statement back into x and y.
This is a general moral worth taking: choose coordinates in which the transformation is simple, do the work there, and translate back. The rectangular formulas look complicated because they are describing a simple operation in an unsuitable language.
Section
Section 2
Concept
To convert an equation, replace every x and y by their expressions in the primed variables and expand. The algebra is heavy but entirely routine, and the reward is an equation in standard form.
The book does not print the full computation and encourages you to do it. That is worth taking seriously once: the mechanism is only convincing after you have watched the cross terms cancel.
Figure (svg): A tilted ellipse whose equation contains a cross term, shown alongside the same ellipse in a rotated frame where the cross term has vanished
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 975-976
Picture it
The same ellipse, seen through two sets of axes.
Figure (svg): A tilted ellipse whose equation contains a cross term, shown alongside the same ellipse in a rotated frame where the cross term has vanished
The curve did not move. Only the frame did, and the cross term was the record of the mismatch between the frame and the curve.
Worked example
Example 11.6.1, part 2. The rotation angle is given.
\[ \text{Convert } 21x^2 + 10xy\sqrt{3} + 31y^2 = 144 \text{ using } \theta = \tfrac{\pi}{3}. \]
Write the substitutions
Why: With cosine pi over 3 equal to one half.
\[ x = x' / 2 - y' \sqrt{3} / 2 \]
Square each and form the cross product
Why: Three expansions to do carefully.
\[ x ^{2}, x y, y ^{2} \]
Substitute and collect
Why: The cross terms cancel.
\[ 36(x') ^{2} + 16(y') ^{2} = 144 \]
Divide to standard form
Why: By 144.
\[ (x') ^{2} / 4 + (y') ^{2} / 9 = 1 \]
Figure (svg): The solution to Worked example a tilted ellipse shown as a ladder of expressions, one row per legal move
\[ \frac{(x')^2}{4} + \frac{(y')^2}{9} = 1 \]
Verify: read off the features
Why: The larger denominator is under y-prime squared, so the major axis lies along the y-prime axis with vertices at 0 comma plus or minus 3, and the minor axis endpoints are at plus or minus 2 comma 0. In the original frame, the major axis is the line through the origin at pi over 3 plus pi over 2.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 975-976
Faded example
With theta equal to pi over 4, both trigonometric values are root two over two.
Fill in the blanks
x = \tfrac-}-(x' - y') \;\Longrightarrow\; x^2 = \tfrac______ ___ x'y' + \tfrac______, \quad xy = \tfrac______ ___ \tfrac______
Why: Squaring the difference gives a negative cross term, and the product xy comes out with no cross term at all, since the two middle contributions cancel. That second fact is why a 45 degree rotation is so clean, and it is exactly what makes the equal-coefficient case easy.
Worked example
Example 11.6.2, part 1. Here the linear terms survive and must be completed.
\[ \text{Graph } 5x^2 + 26xy + 5y^2 - 16x\sqrt{2} + 16y\sqrt{2} - 104 = 0. \]
Find the angle
Why: A and C are equal so the cotangent is zero.
\[ \theta = \frac{\pi}{4} \]
Substitute
Why: Both trigonometric values are root two over two.
Collect
Why: The cross term cancels; a linear term in y-prime survives.
\[ 18(x') ^{2} - 8(y') ^{2} + 32 y' - 104 = 0 \]
Complete the square in y-prime
Why: And divide to standard form.
\[ (x') ^{2} / 4 - (y' - 2) ^{2} / 9 = 1 \]
Figure (svg): The solution to Worked example a tilted hyperbola shown as a ladder of expressions, one row per legal move
\[ \frac{(x')^2}{4} - \frac{(y'-2)^2}{9} = 1 \]
Verify: check the features
Why: The positive term is in x-prime, so the hyperbola opens in the x-prime direction with vertices at plus or minus 2 comma 2 and asymptotes of slope plus or minus three halves through the centre. Note both original squared coefficients were positive, and the curve is nonetheless a hyperbola.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 977-978
Error analysis
A student substitutes into a squared term.
Annotate
On: \( x = \tfrac{x'}{2} - \tfrac{y'\sqrt{3}}{2} \;\Longrightarrow\; x^2 = \tfrac{(x')^2}{4} - \tfrac{3(y')^2}{4} \)
The cross terms from each squared substitution are the whole point of the exercise, since their cancellation is what removes the x-prime y-prime term. Dropping them removes the mechanism along with the labour.
Prediction
An equation has A equal to C, so its two squared coefficients agree.
Predict first
What rotation angle removes the cross term?
Correct: pi/4.
Why: The condition is that the cotangent of twice theta equals A minus C over B, and when A equals C the numerator is zero. So the cotangent of twice theta is zero, giving twice theta equal to pi over 2 and theta equal to pi over 4. This is a case worth recognising by sight, since it makes all the subsequent arithmetic clean.
Sorting
Rotation changes some features of a conic and preserves others.
Sort into buckets
Sort each property.
Edge cases
The method assumes an angle exists that removes the cross term.
Discussion prompt
What happens if B is already zero, and does the theorem still say anything?
Hint: What does the condition become?
Answer:
If B is zero there is nothing to remove — the equation is already in a frame aligned with the conic's axes, and the standard forms from earlier apply directly.
The theorem explicitly assumes B is not zero, because the condition involves dividing by B. That exclusion is a genuine one rather than a formality.
Geometrically it means the conic's axes are already parallel to the coordinate axes, so the correct rotation is by zero. The cross term is precisely the measure of how far the frame is from the curve's own axes, and B equal to zero says the mismatch is already gone.
Section
Section 3
Concept
Substitute the rotation formulas into the general second-degree equation, collect the coefficient of the cross term, and set it to zero. Two double-angle identities reduce the result to a single condition.
Theorem 11.10 — The equation with a non-zero cross-term coefficient B can be transformed into one with no cross term by rotating counter-clockwise through any angle theta satisfying the cotangent condition.
\[ \cot(2\theta) = \frac{A - C}{B} \]
The justification for the division is the same Pythagorean argument used twice already in this chapter, and it is worth recognising as a recurring move rather than a new trick.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 976-977
Picture it
What you actually need is the cosine and sine, not the angle itself.
Figure (svg): The two routes from the cotangent condition to the sine and cosine of the rotation angle: directly via the arccotangent and half-angle identities, or via the double-angle identity for tangent
Route two turns the condition into a quadratic in the tangent of theta, whose acute root builds a right triangle from which both values are read directly.
Worked example
Collecting the cross-term coefficient and setting it to zero.
\[ \text{Show the cross term vanishes when } \cot(2\theta) = \tfrac{A-C}{B}. \]
Collect the cross-term contributions
Why: From the three quadratic terms.
Apply the double angle identities
Why: Twice cosine sine is the sine of twice the angle.
Set to zero and rearrange
Why: Group the sine terms.
Divide, after justifying
Why: By B and by the sine of twice theta.
Figure (svg): The solution to Worked example the derivation of the condition shown as a ladder of expressions, one row per legal move
\[ \cot(2\theta) = \frac{A - C}{B} \]
Verify: justify the division
Why: If the sine of twice theta were zero, the equation would force B times the cosine of twice theta to be zero, and since B is non-zero the cosine would have to vanish too. No angle has both, so the sine of twice theta is non-zero on the solution set and the division is safe.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 976-977
Faded example
For 3x squared plus 4xy plus 3y squared equal to 5, find the rotation angle.
Fill in the blanks
\cot(2\theta) = \frac0pi/4 = ___ \;\Longrightarrow\; 2\theta = \tfrac______ \;\Longrightarrow\; \theta = ___
Why: Equal squared coefficients make the numerator zero, so the cotangent of twice theta is zero and twice theta is a right angle. This case is worth spotting immediately, since it is common and produces the cleanest possible substitutions.
Worked example
Example 11.6.2, part 2. The book's preferred route.
\[ \text{For } 16x^2 + 24xy + 9y^2 + 15x - 20y = 0, \text{ find } \cos\theta \text{ and } \sin\theta. \]
Compute the cotangent
Why: A minus C over B.
Invert and use the double angle for tangent
Why: Twice the tangent over one minus its square.
\[ 24 \tan ^{2} + 14 \tan - 24 = 0 \]
Factor and take the acute root
Why: Two factors, one positive root.
\[ \tan \theta = \frac{3}{4} \]
Build the right triangle
Why: Legs 3 and 4, hypotenuse 5.
\[ \cos = \frac{4}{5}, \sin = \frac{3}{5} \]
Figure (svg): The solution to Worked example an angle that is not standard shown as a ladder of expressions, one row per legal move
\[ \tan\theta = \tfrac{3}{4} \;\Longrightarrow\; \cos\theta = \tfrac{4}{5}, \; \sin\theta = \tfrac{3}{5} \]
Verify: finish the problem
Why: Substituting these gives 25 times x-prime squared minus 25 y-prime equal to zero, so y-prime equals x-prime squared — a parabola with vertex at the origin opening along the positive y-prime axis. A clean result from an unpromising equation.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 978-978
Trap
\[ 2(3\tan\theta + 4)(4\tan\theta - 3) = 0 \;\Longrightarrow\; \tan\theta = -\tfrac{4}{3} \]
Take either root, since both satisfy the original condition
Why: Both genuinely do satisfy the cotangent equation.
But the negative root gives an obtuse angle, which rotates the axes further than necessary and produces a correct but awkward frame with the roles of the two axes swapped.
\[ \tan\theta = \tfrac{3}{4} \;\Longrightarrow\; \theta = \arctan\left(\tfrac{3}{4}\right), \text{ acute} \]
Choose the acute angle
Why: It is always available and it keeps the frame in the natural orientation.
Both answers are legitimate rotations that remove the cross term. Choosing the acute one is a convention, adopted so that different people solving the same problem produce comparable frames — the same reason the standard polar range exists.
Prediction
You need the cosine and sine of the rotation angle, and the cotangent of twice theta is not a common value.
Predict first
Which route is less work?
Correct: Double angle for tangent, then a right triangle.
Why: Turning the condition into a quadratic in the tangent of theta gives an exact rational value for that tangent, from which a right triangle supplies exact values for the cosine and sine. The arccotangent route obtains theta itself and then needs half-angle identities with their sign decisions to reach the same two numbers. The book adopts the second route for exactly this reason, and no decimal approximation is needed by either.
Matching
The cotangent condition decides.
Match the pairs
Why: The first has equal squared coefficients, giving the standard 45 degree case. The second gives negative ten over ten root three, which simplifies to the cotangent of two thirds of pi and so theta is pi over 3. The third gives a value requiring inverse functions. The fourth has no cross term at all, so the theorem does not apply and no rotation is wanted.
Explain it to yourself
The condition involves the cotangent of twice theta rather than of theta.
Discussion prompt
Explain where the doubling comes from, and why that is natural rather than accidental.
Hint: Which identities were used in the derivation?
Answer:
The doubling comes from the double angle identities. The cross-term coefficient contains products of a cosine and a sine of theta, and twice such a product is the sine of twice theta; it also contains the difference of their squares, which is the cosine of twice theta.
So the whole coefficient is naturally an expression in twice theta, and the condition it produces is about twice theta too.
This is not accidental. A rotation by theta and a rotation by theta plus pi produce the same pair of axes, just relabelled, so any condition on the axes can only see theta up to a half turn — which is to say, it can only see twice theta. The doubling in the formula is the algebra reflecting that geometric fact, and it also explains why the condition has more than one acute-looking solution.
Section
Section 4
Concept
The old rule of thumb — read the conic off the signs of the squared coefficients — fails completely once a cross term is present. Both worked examples have positive coefficients, and one is a hyperbola and the other a parabola.
This is a genuine loss. Everything learned about reading a conic off its equation applies only when the frame happens to be aligned with the curve, and a cross term is the announcement that it is not.
Figure (svg): Two columns contrasting what the coefficients of the squared terms tell you when there is no cross term against when there is one
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 977-979
Picture it
The left column is what you knew. The right column is what a cross term costs.
Figure (svg): Two columns contrasting what the coefficients of the squared terms tell you when there is no cross term against when there is one
Every instinct built on the left column has to be suspended once B is non-zero, which is the practical reason the discriminant matters.
Worked example
Example 11.6.2, part 1, completed and read.
\[ \text{Read the features of } \frac{(x')^2}{4} - \frac{(y'-2)^2}{9} = 1. \]
Identify the centre
Why: In the primed frame.
\[ (0, 2) \]
Identify the opening direction
Why: The positive term is in x-prime.
Find the vertices
Why: Two units either side of the centre.
\[ (+- 2, 2) \]
Find the asymptote slopes
Why: Three over two.
\[ \text{slope } +- \frac{3}{2} \]
Figure (svg): The solution to Worked example the hyperbola case in full shown as a ladder of expressions, one row per legal move
\[ \text{centre } (0, 2), \; \text{vertices } (\pm 2, 2), \; y' = \pm\tfrac{3}{2}x' + 2 \]
Verify: state the frame explicitly
Why: Every one of those coordinates is in the primed frame, which is the original rotated by 45 degrees. Reporting them without saying so would be wrong, since the centre is not at 0 comma 2 in the original coordinates. Naming the frame is part of the answer.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 977-978
Two truths and a lie
Three of these are true and one is false.
Eliminate the wrong options
One of these statements about conics with a cross term is wrong.
Survives elimination: B
Why: Statement B is false, and both worked examples of this lesson refute it. The equation with coefficients 5 and 5 is a hyperbola and the one with 16 and 9 is a parabola, though all four numbers are positive. The rule holds only when the cross-term coefficient is zero.
Worked example
Example 11.6.2, part 2. An unpromising equation with a very clean answer.
\[ \text{Complete the reduction of } 16x^2 + 24xy + 9y^2 + 15x - 20y = 0. \]
Use the values found earlier
Why: Cosine four fifths and sine three fifths.
\[ x = \frac{4 x' - 3 y'}{5} \]
Substitute and expand
Why: All three quadratic expansions.
Collect
Why: The quadratic terms collapse to one.
\[ 25(x') ^{2} - 25 y' = 0 \]
Divide
Why: By 25.
\[ y' = (x') ^{2} \]
Figure (svg): The solution to Worked example the parabola case in full shown as a ladder of expressions, one row per legal move
\[ y' = (x')^2 \]
Verify: check the classification independently
Why: The discriminant is 24 squared minus 4 times 16 times 9, which is 576 minus 576, exactly zero. Zero discriminant means a parabola, agreeing with the reduction — and available before any of the algebra was done.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 978-978
Error analysis
A student looks at the equation before rotating.
Annotate
On: \( 5x^2 + 26xy + 5y^2 - 16x\sqrt{2} + 16y\sqrt{2} - 104 = 0 \;\Longrightarrow\; \text{both positive, so an ellipse} \)
The signs of A and C describe the equation in the frame it happens to be written in. When that frame is tilted relative to the curve, they describe nothing about the curve at all.
Prediction
A rotation reduces an equation to a standard form with a centre at 0 comma 2.
Predict first
Where is the centre in the original coordinates?
Correct: At the point (0, 2) converted back through the rotation.
Why: The coordinates 0 comma 2 describe the centre in the rotated frame. Converting back requires the new-to-old system: x equals 0 times cosine minus 2 times sine, and y equals 0 times sine plus 2 times cosine. Reporting 0 comma 2 as though it were an original coordinate is a common and serious slip, and naming the frame in every answer prevents it.
Sorting
Rotation produces answers in the primed frame unless converted back.
Sort into buckets
Sort each quantity by whether it needs converting before being reported in the original frame.
Counterexample
A student proposes: an equation whose squared terms both have positive coefficients cannot describe a hyperbola.
Discussion prompt
Give a counterexample, and say precisely when their rule is safe.
Hint: Try the simplest possible cross term.
Answer:
Take x squared plus 4xy plus y squared equal to 1. Both squared coefficients are positive, but the discriminant is 16 minus 4, namely 12, which is positive — so the curve is a hyperbola. Rotating by 45 degrees turns it into 3 times x-prime squared minus y-prime squared equal to 1, which is unmistakably one.
The rule is safe exactly when the cross-term coefficient is zero. Then the equation is already in a frame aligned with the curve's axes, and the signs of the squared coefficients genuinely describe the curve.
The general moral is worth stating plainly: a rule read off an equation is really a rule about the frame the equation is written in. When the frame changes, the rule may say nothing at all, and the discriminant exists precisely because it is one quantity that does not care which frame you chose.
Section
Section 5
Concept
The quantity B squared minus 4AC classifies any non-degenerate conic, and it does so without any rotation being performed. Its sign is the opposite of the sign of the product of the rotated squared coefficients.
Theorem 11.11 — For a non-degenerate conic: a positive discriminant gives a hyperbola, a zero discriminant gives a parabola, and a negative discriminant gives an ellipse or circle.
\[ B^2 - 4AC = -4A'C' \]
The proof runs on the power reduction formulas and the cotangent condition, and every trigonometric term cancels by the end. That the answer contains no theta at all is the point: the discriminant does not depend on the frame.
Figure (svg): The three cases of the discriminant of a general second-degree equation, each labelled with the conic it identifies
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 979-981
Picture it
One computation, three outcomes, no rotation.
Figure (svg): The three cases of the discriminant of a general second-degree equation, each labelled with the conic it identifies
Running this first costs one line and tells you what to expect from a page of algebra, which makes it a check as well as a shortcut.
Worked example
Example 11.6.3. The three equations from the lesson, classified in one line each.
\[ \text{Classify the three worked equations by their discriminants.} \]
First equation
Why: A is 21, B is ten root three, C is 31.
\[ 300 - 2604 = -2304 \]
Second equation
Why: A is 5, B is 26, C is 5.
\[ 676 - 100 = 576 \]
Third equation
Why: A is 16, B is 24, C is 9.
\[ 576 - 576 = 0 \]
Read the signs
Why: Negative, positive, zero.
Figure (svg): The solution to Worked example classifying three conics shown as a ladder of expressions, one row per legal move
\[ -2304 < 0: \text{ ellipse}; \quad 576 > 0: \text{ hyperbola}; \quad 0: \text{ parabola} \]
Verify: compare against the reductions
Why: The first reduced to an ellipse in standard form, the second to a hyperbola, and the third to a parabola. All three agree with the discriminant, which took one line each against a page each.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 981-981
Faded example
Classify the conic 2x squared plus 5xy plus 2y squared minus 7 equal to 0.
Fill in the blanks
B^2 - 4AC = 25 - 4(2)(2) = 9, \texthyperbola ___
Why: The discriminant is 25 minus 16, namely 9, which is positive. So the curve is a hyperbola — despite both squared coefficients being positive and equal, which under the old rule would have suggested a circle.
Worked example
The key line of the proof, in outline.
\[ \text{Explain why } B^2 - 4AC \text{ has the opposite sign to } A'C'. \]
Write the rotated coefficients
Why: Using the power reduction formulas.
\[ 2 A'\text{ and } 2 C' \]
Form their product
Why: A difference of squares plus cross terms.
\[ 4 A' C' \]
Substitute the cotangent condition
Why: Twice, together with the Pythagorean Identity.
Read the result
Why: The product is 4AC minus B squared.
\[ 4 A' C' = 4 AC - B ^{2} \]
Figure (svg): The solution to Worked example why the sign flips shown as a ladder of expressions, one row per legal move
\[ 4A'C' = 4AC - B^2 \;\Longrightarrow\; B^2 - 4AC = -4A'C' \]
Verify: check against a known case
Why: For the hyperbola, the rotated coefficients were 18 and negative 8, whose product is negative 144, so negative four times that is 576. The discriminant computed directly was also 576. The identity holds.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 979-980
Trap
\[ x^2 + 2xy + y^2 = 0 \;\Longrightarrow\; B^2 - 4AC = 4 - 4 = 0 \;\Longrightarrow\; \text{a parabola} \]
Compute the discriminant and read off the conic
Why: The computation is correct and the discriminant is genuinely zero.
But the equation factors as x plus y all squared equal to zero, so its graph is the single line y equals negative x. That is a degenerate case, and the theorem explicitly excludes them.
Check for degeneracy first, by looking for a factorisation or for an empty solution set.
The theorem's hypothesis is that the equation describes a non-degenerate conic — a genuine circle, parabola, ellipse or hyperbola. Degenerate cases give points, lines, pairs of lines, or nothing.
This is why the hypothesis is stated rather than assumed. A discriminant tells you which family a genuine conic belongs to; it does not tell you that you have one.
Sorting
Compute the discriminant and read the sign.
Sort into buckets
Sort each equation.
Prediction
You rotate a conic through the angle that removes the cross term, then compute the discriminant of the new equation.
Predict first
How does it compare with the original discriminant?
Correct: It is the same number.
Why: With the cross term gone, the new discriminant is 0 minus 4 A-prime C-prime, and the proof of the theorem shows that this equals B squared minus 4AC exactly. The discriminant is unchanged by rotation, which is precisely why it can classify the conic without one. A quantity that survives the transformation is what any such shortcut has to be built from.
Real world
A structural engineer analysing stress at a point in a loaded beam gets a two by two array of numbers: two normal stresses and a shear stress. Rotating the coordinate frame changes all three, and there is one orientation, called the principal axes, at which the shear vanishes.
Discussion prompt
Explain the correspondence with this lesson, and say what the discriminant becomes.
Hint: Which of the three numbers plays the role of the cross-term coefficient?
Answer:
The shear stress plays the role of B, and the two normal stresses are A and C. Finding the principal axes is finding the rotation that makes the shear zero, and the angle is given by exactly the same cotangent condition — engineers write it as a tangent of twice theta, which is the same equation rearranged.
The discriminant becomes a frame-independent measure of the stress state, one of the quantities engineers call invariants. Whether it is positive, zero or negative distinguishes genuinely different physical situations, in the same way it distinguishes hyperbola from parabola from ellipse.
The construction is drawn as Mohr's circle, and its geometry is the doubling in the cotangent of twice theta made visible: a rotation of the physical axes by theta moves the point on the circle by 2 theta. Every engineering student meets it, usually without being told it is the same theorem as this one.
The general point worth carrying: the interesting quantities are the ones that survive a change of frame. The coefficients A, B and C describe the equation you happened to write; the discriminant describes the situation itself, and that is why both a mathematician and an engineer end up computing it.
Comparison
Fill the blanks from memory. The last row is the one that makes the rest optional.
Comparison matrix
| Before rotating | After rotating | |
|---|---|---|
| the equation | has a cross term | no cross term |
| reading the conic off the signs | unreliable | reliable, as in earlier chapters |
| the curve itself | unchanged | unchanged |
| the discriminant | the same number either way | now equal to -4 A' C' |
The third and fourth rows say the same thing from two directions. The curve does not care which frame you use, so any quantity describing the curve rather than the equation must be unchanged by rotation.
Pattern
Five moves, and the first one may make the rest unnecessary.
The discriminant is unchanged by rotation, so it can be recomputed after step 4 as a check.
OpenStax Algebra and Trigonometry 2e, §12.4 Rotation of Axes §12.4
Check
The rotation angle.
Check your understanding
For the equation 7x squared plus 6xy minus y squared equal to 4, what does the cotangent of twice the rotation angle equal?
Answer: A
Why: The condition is A minus C over B, with A equal to 7, B equal to 6 and C equal to negative 1. So the numerator is 7 minus negative 1, which is 8, and the quotient is 8 over 6, namely four thirds.
Check
The discriminant.
Check your understanding
Classify the conic 4x squared minus 4xy plus y squared plus 3x minus 2 equal to 0.
Answer: B
Why: The discriminant is negative 4 squared minus 4 times 4 times 1, which is 16 minus 16, exactly zero. A zero discriminant identifies a parabola, and the presence of a linear term rules out the degenerate case of a repeated line.
Check
Frames and answers.
Check your understanding
A rotation by 30 degrees reduces a conic to a standard ellipse with vertices at (0, ±3) in the primed frame. What is true of the original curve?
Answer: B
Why: The length 6 is intrinsic and survives the rotation. The direction does not: the major axis lies along the y-prime axis, which is the y-axis rotated by 30 degrees, and so points at 90 plus 30, namely 120 degrees.
Real world
A machine-learning practitioner has a cloud of two-dimensional data points that are strongly correlated: the cloud is a tilted elliptical blob. They want to describe it with two uncorrelated coordinates, one along the direction of greatest spread and one perpendicular to it.
Discussion prompt
Explain what they are doing in the language of this lesson, and what the cross term corresponds to.
Hint: What quantity are they trying to make zero?
Answer:
The cloud's shape is captured by a quadratic form in x and y, and the cross term is the correlation. Making it zero is exactly the problem of this lesson, and the angle that does it is found from the same condition.
The rotated axes are called the principal components, and the practitioner then keeps the one along the direction of greatest spread and often discards the other. That is dimensionality reduction, and it is a rotation of axes followed by a projection.
The eigenvalues of the associated matrix are the rotated coefficients A-prime and C-prime, and their product is what the discriminant measures. A near-zero smaller eigenvalue means a nearly degenerate blob — a cloud that is effectively one-dimensional, which is exactly the condition that makes the reduction worthwhile.
The same mathematics appears again as the moment of inertia tensor in mechanics, as stress in a beam, and as the covariance matrix in statistics. All four are asking for the frame in which a symmetric two-index quantity becomes diagonal, and all four solve it by rotating until the cross term dies.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
An equation has A and C both positive and a large positive B. What can you conclude about the conic?
Correct: Nothing yet; compute the discriminant.
\[ 5x^2 + 26xy + 5y^2 - \cdots = 0: \quad B^2 - 4AC = 576 > 0 \;\Longrightarrow\; \text{hyperbola} \]
Why: With a cross term present, the signs of the squared coefficients carry no information about the conic. A large B makes B squared large, which pushes the discriminant positive and does suggest a hyperbola, but suggest is not conclude — the answer depends on the size of 4AC as well. One subtraction settles it, and nothing short of that does.
Explain it
They have met the standard conic forms and are confused by an equation with an xy term, which matches none of them.
Discussion prompt
In no more than five sentences, explain what the xy term means and what to do about it.
Hint: What is the term telling you about the axes rather than about the curve?
Answer:
The xy term means the curve is tilted relative to your axes. It is not a different kind of curve — it is one of the same four, looked at from an inconvenient angle.
Every standard form assumes the curve's own axes line up with the coordinate axes. The cross term is the measure of how far off that alignment is, and it disappears exactly when the two agree.
So rotate the axes until it does. The angle comes from the cotangent of twice it being A minus C over B, and after substituting, the equation lands in a standard form you already know how to read.
If you only need to know which conic it is, skip all of that and compute B squared minus 4AC. Positive is a hyperbola, zero a parabola, negative an ellipse.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Rotating a point is fixed by knowing which of the two systems produces the coordinates you want. Finding the angle is fixed by the cotangent condition plus the double-angle route to the cosine and sine. The substitution is fixed by expanding every binomial square in full, cross terms included. Classifying is fixed by one subtraction. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page draw two sets of axes sharing an origin, one rotated from the other, with a point marked and both its angles labelled, and write the four-line derivation of the rotation equation for x beside it. Underneath, write both systems of Theorem 11.9 and note in one line which system converts equations and which converts points. In the middle of the page write the cotangent condition, and beside it the two routes to the cosine and sine of the rotation angle, marking which one the book prefers and why. In the bottom left, take the equation with coefficients 5, 26 and 5, find the angle, and state the standard form it reduces to along with the frame that form is in. In the bottom right, write the three cases of the discriminant, and apply it to all three equations from this lesson. Finally, circle the one quantity on the page that does not change when the axes rotate.
The circled quantity is the discriminant. That it is unchanged by rotation is exactly why it can classify a conic without one being performed, and it is why the proof of Theorem 11.11 ends with every trigonometric term cancelled away.
Recap
Five things, and the fifth makes the other four optional whenever only the classification is wanted.
| If the question says | Your first move is |
|---|---|
| Which conic is this? | Compute B squared minus 4AC |
| Graph this equation with a cross term | Find the angle from the cotangent condition |
| The cotangent is not a common value | Use the double angle for tangent, then a right triangle |
| Convert this point | Choose the system giving the coordinates you want |
| Report the centre or vertices | Say which frame they are measured in |
The conics have now been rotated back into shapes you can read. The next lesson takes the opposite approach: rather than repairing the rectangular equation, it redefines all four conics at once by a single ratio, and finds them a polar form in which the eccentricity is visible at a glance.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.6 Hooked on Conics Again §11.6, pp. 973-981 — everything on these slides traces back here
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