Explains the cross term as a sign that a conic is tilted, gives the rotation formulas and the angle that removes it, and introduces the discriminant — a quantity unchanged by rotation that classifies any second-degree equation without rotating it at all.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 10 — Analytic Geometry
§10.4 Rotation of Axes, pp. 1248-1265
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1248-1265 — the pages these objectives are drawn from
Warm-up
Every conic so far had squared terms and linear terms, and nothing else.
Discussion prompt
What could the presence of an xy term in a second-degree equation indicate?
Hint: What kind of change would introduce it?
Answer:
Substituting a rotation into a standard conic equation produces one. Each rotated coordinate mixes both original ones, so squaring produces a cross term.
So a cross term means the conic is tilted relative to the axes — not that it is a different kind of curve.
Which suggests the remedy: rotate the axes to line up with the conic's own directions, and the cross term disappears. That is the whole technique of the section.
Concept
The general second-degree equation describes a conic in any orientation. A cross term appears exactly when the conic's axes are not parallel to the coordinate axes.
\[ Ax^2+Bxy+Cy^2+Dx+Ey+F=0 \]
The cross term's coefficient measures the tilt, and choosing the right rotation angle makes it zero. After that the equation is one of the standard forms from the previous three sections.
Figure (svg): An ellipse drawn tilted relative to the coordinate axes, with a second pair of axes drawn along its own directions
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1248-1252
Section
Section 1
Concept
The general second-degree equation in two variables covers every conic in every position and orientation, with the cross term's coefficient carrying the tilt.
Reading the equation's structure tells you what to do before any computation: linear terms mean complete the square, a cross term means rotate first. Doing them in the wrong order makes the algebra much worse.
Figure (svg): An ellipse drawn tilted relative to the coordinate axes, with a second pair of axes drawn along its own directions
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1248-1253
Picture it
Its own axes, dashed, are not the coordinate axes.
Figure (svg): An ellipse drawn tilted relative to the coordinate axes, with a second pair of axes drawn along its own directions
In the dashed frame the ellipse's equation is one of §10.1's standard forms. The cross term exists only because the equation is written in the wrong frame for this curve.
Worked example
Which terms are present says what to do.
\[ \text{What does } 5x^2-4xy+2y^2-30=0 \text{ tell you before any work?} \]
Check for a cross term
Why: Present.
Check for linear terms
Why: Absent.
Conclude the position
Why: No translation needed.
Note the plan
Why: Rotate to remove the cross term.
Figure (svg): An ellipse drawn tilted relative to the coordinate axes, with a second pair of axes drawn along its own directions
\[ \text{rotated, not translated} \]
Verify: confirm the reading
Why: The absence of linear terms means substituting the negative of any point gives the same equation, so the curve is symmetric about the origin — which is exactly what being centred there means. The cross term is the only complication.
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1249-1251
Sorting
Each kind of term means something about position.
Sort into buckets
Sort each term.
Worked example
Rotating a standard conic produces one.
\[ \text{Substitute a rotation into } X^2-Y^2=1 \text{ and see what appears.} \]
Write the substitution
Why: Each new coordinate mixes both old.
Square each
Why: Cross products appear.
Subtract
Why: The cross terms do not cancel.
Conclude
Why: Rotation introduces the cross term.
Figure (svg): The solution to Worked example recover a cross term shown as a ladder of expressions, one row per legal move
\[ \text{a cross term appears} \]
Verify: check a special angle
Why: Rotating by 45 degrees turns this hyperbola into the familiar reciprocal curve, whose equation is a product of the two variables equalling a constant — a pure cross term with no squared terms at all. That is the extreme case of the same phenomenon.
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1251-1253
Trap
\[ \text{there are linear terms, so complete the square first} \]
Handle the translation before the rotation
Why: The usual first move is applied without checking for a cross term.
The cross term makes the squares impossible to complete cleanly, and the work has to be redone.
Rotate first, then complete the square. The cross term prevents the squares from separating.
After the rotation the equation has no cross term and completing the square works exactly as in the previous sections.
Reading the structure first settles the order: a cross term means rotation comes before anything else.
Prediction
A general equation has a cross term but no linear terms.
Predict first
What follows?
Correct: The conic is centred at the origin but rotated.
Why: Linear terms record a translation and the cross term records a rotation. With one present and the other absent, only a rotation is needed to reach standard form.
Faded example
From the equation with squared coefficients 5 and 2 and cross coefficient negative 4.
Fill in the blanks
A=5, \quad B=-4, \quad C=2
Why: The three coefficients of the second-degree terms are what both the discriminant and the rotation angle depend on. Reading them off correctly, signs included, is the first step of either computation.
Step zero
You are given a general second-degree equation.
Discussion prompt
What do you check before doing anything?
Hint: Which term changes the plan?
Answer:
Whether there is a cross term. If there is, the conic is tilted and a rotation must come before anything else.
Completing the square first does not work with a cross term present, because the squares cannot be separated while the variables are mixed.
And if only the type is wanted, the discriminant answers that without any rotation at all. Checking for the cross term decides between three quite different amounts of work.
Section
Section 2
Concept
The quantity formed from the three second-degree coefficients is unchanged by rotation, so its sign identifies the conic type directly from the tilted equation.
Being unchanged by rotation is what makes it useful. A quantity that changed would have to be computed after rotating, which would defeat the purpose of having a shortcut.
| discriminant | conic type |
|---|---|
| negative | an ellipse, or a circle |
| zero | a parabola |
| positive | a hyperbola |
Figure (svg): A card giving the discriminant and the three conic types it distinguishes by sign
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1253-1257
Picture it
One computation, three outcomes.
Figure (svg): A card giving the discriminant and the three conic types it distinguishes by sign
The red line is why it matters. Because the value survives rotation, the tilted equation can be classified as easily as an aligned one, with no rotation performed.
Worked example
One computation.
\[ \text{Classify } 5x^2-4xy+2y^2-30=0. \]
Read the coefficients
Why: Signs included.
\[ A = 5, B = -4, C = 2 \]
Square the cross coefficient
Why: Sign disappears.
\[ 16 \]
Compute four times the product
Why: Of the two squared coefficients.
\[ 40 \]
Subtract
Why: The discriminant.
\[ 16 - 40 = -24 \]
Figure (svg): A card giving the discriminant and the three conic types it distinguishes by sign
\[ -24<0: \text{an ellipse} \]
Verify: sanity check the sign
Why: Both squared coefficients are positive here, which makes the subtracted term large and the discriminant negative — consistent with an ellipse. A hyperbola would need the squared coefficients to have opposite signs or the cross term to dominate.
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1254-1256
Prediction
The discriminant comes out positive.
Predict first
Which conic is it?
Correct: A hyperbola.
Why: A positive discriminant identifies the hyperbola in any orientation. Negative gives an ellipse and zero a parabola, and the value is unchanged by rotation so the tilted equation can be used directly.
Worked example
The discriminant vanishes.
\[ \text{Classify } x^2-4xy+4y^2+5x-10=0. \]
Read the coefficients
Why: From the second-degree terms.
\[ A = 1, B = -4, C = 4 \]
Square the cross coefficient
Why: Sixteen.
\[ 16 \]
Compute four times the product
Why: Also sixteen.
\[ 16 \]
Subtract
Why: Zero.
Figure (svg): The solution to Worked example a parabola shown as a ladder of expressions, one row per legal move
\[ 0: \text{a parabola} \]
Verify: look at the squared terms
Why: The three second-degree terms form a perfect square, which is exactly what a zero discriminant detects — the quadratic part factors into a repeated linear factor. That is the algebraic signature of a parabola in any orientation.
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1256-1257
Error analysis
A student computes a discriminant.
Annotate
On: \( A=5,\;B=-4,\;C=2 \;\Longrightarrow\; B^2-4AC=-16-40=-56 \)
Squaring the cross coefficient removes its sign, so a negative cross term contributes positively. The error is harmless in some cases and decisive in others, which makes it worth avoiding rather than hoping it does not matter.
Faded example
With coefficients 5, negative 4 and 2.
Fill in the blanks
(-4)^2-4(5)(2)=16-40=-24
Why: Squaring the cross coefficient makes it positive regardless of its sign, and four times the product of the squared coefficients is subtracted. The sign of the result is all that matters for the classification.
Sorting
The sign decides.
Sort into buckets
Sort each value.
Explain it to yourself
The discriminant is unchanged by rotating the axes.
Discussion prompt
Explain why that property is what makes it useful.
Hint: What would a changing quantity require?
Answer:
The three second-degree coefficients all change under rotation — that is the whole point of rotating, since the cross coefficient is driven to zero.
But this particular combination of them comes out the same in every frame. So it can be computed before the rotation and still describes the rotated equation.
A quantity that changed would have to be evaluated after rotating, which would mean doing the work the shortcut was meant to avoid. A good explanation notes that invariance is exactly what makes a shortcut possible, not an incidental property.
Section
Section 3
Concept
Rotating the axes expresses each original coordinate as a combination of the new ones, weighted by the cosine and sine of the rotation angle.
It is the axes that rotate, not the curve. A point keeps its position and acquires new coordinates because it is being described from a different frame — which is why the conic's shape and size are unchanged by the substitution.
Figure (svg): Three cards giving the rotation formulas, the angle that removes the cross term, and the special case when the two squared coefficients are equal
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1257-1260
Picture it
Substitution, angle, and the special case.
Figure (svg): Three cards giving the rotation formulas, the angle that removes the cross term, and the special case when the two squared coefficients are equal
The second card is where the angle comes from. It is not chosen for convenience but computed from the coefficients, as the unique value that makes the cross term vanish.
Worked example
Substitute and expand.
\[ \text{Rotate } xy=1 \text{ by } 45^\circ. \]
Write the substitution
Why: Both cosine and sine are root two over two.
Form the product
Why: Multiply the two expressions.
\[ (X - Y) (X + Y) / 2 \]
Expand
Why: A difference of squares.
\[ \frac{X ^{2} - Y ^{2}}{2} \]
Set equal to one
Why: The original right side.
\[ X ^{2} - Y ^{2} = 2 \]
Figure (svg): Three cards giving the rotation formulas, the angle that removes the cross term, and the special case when the two squared coefficients are equal
\[ X^2-Y^2=2 \]
Verify: check the discriminant both ways
Why: The original has coefficients 0, 1, 0, giving a discriminant of 1 — positive, a hyperbola. The rotated form has coefficients 1, 0, negative 1, giving 0 minus four times negative one, which is 4 — also positive. Different values are not expected here, so this confirms the invariance.
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1258-1259
Prediction
The axes are rotated by some angle.
Predict first
What changes?
Correct: The coordinates describing each point.
Why: A rotation of axes is a change of description, not of object. Every point stays where it is and acquires new coordinates measured from the new frame, which is why the discriminant and the conic's dimensions are unchanged.
Worked example
Only the description changes.
\[ \text{Does rotating the axes change the conic itself?} \]
Consider a point on the curve
Why: It stays where it is.
Consider its coordinates
Why: Measured from new axes.
Consider the shape
Why: Lengths and angles unchanged.
Conclude
Why: A change of description.
Figure (svg): The solution to Worked example the curve does not move shown as a ladder of expressions, one row per legal move
\[ \text{same curve, new frame} \]
Verify: check an invariant
Why: The discriminant is the same before and after, as are the conic's actual dimensions. Quantities that describe the curve itself are unchanged, and only those describing its relationship to the axes differ — which is exactly what a change of frame should do.
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1259-1260
Trap
\[ \text{rotate the ellipse by }45^\circ\text{ to line it up} \]
Move the curve to match the axes
Why: The conic is treated as the thing being turned.
A different curve results, in a different place, rather than a new description of the same one.
The axes rotate and the curve stays put. The substitution re-describes the same points from a new frame.
That is why the conic's dimensions and the discriminant are unchanged — nothing about the object has been altered.
Rotating the curve would be a transformation of the kind in §1.5, and it would produce a genuinely different set of points.
Faded example
At 45 degrees, where both the cosine and the sine are root two over two.
Fill in the blanks
x=\frac2}2(X-Y), \quad y=\frac___}}}___}(X+Y)
Why: At 45 degrees both trigonometric values are equal, which is why this angle produces the cleanest substitutions. The minus sign appears in one formula and not the other, which is what makes the rotation a rotation rather than a reflection.
Sorting
Descriptions change and objects do not.
Sort into buckets
Sort each quantity.
Explain it
The section is called rotation of axes.
Discussion prompt
Explain to a classmate what moves and what does not.
Hint: Where do the points go?
Answer:
The axes rotate; the curve stays exactly where it is. Every point keeps its position in the plane.
What changes is how each point is described — its coordinates are now measured from the new axes, so the numbers differ even though the point does not.
That is why the conic's size, shape and type are unchanged. A good explanation contrasts this with §1.5's transformations, which genuinely move the curve and would produce a different set of points.
Section
Section 4
Concept
The rotation angle is determined by the three second-degree coefficients. It is the unique choice, up to right angles, that makes the new cross coefficient zero.
That several angles work is expected: rotating a further right angle swaps which axis the conic opens along without introducing a cross term. Any of them reaches a standard form, so the smallest positive one is conventional.
Figure (svg): Three cards giving the rotation formulas, the angle that removes the cross term, and the special case when the two squared coefficients are equal
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1260-1263
Picture it
The second card, with the equal-coefficient case beside it.
Figure (svg): Three cards giving the rotation formulas, the angle that removes the cross term, and the special case when the two squared coefficients are equal
The equal case is worth recognising on sight, since the cotangent is then zero and the angle is 45 degrees without any computation.
Worked example
A cotangent of zero.
\[ \text{Find the rotation angle for } 3x^2+4xy+3y^2-10=0. \]
Compare the squared coefficients
Why: Both three.
Compute the cotangent
Why: Difference over the cross coefficient.
\[ 0 \]
Solve for twice the angle
Why: Where the cotangent vanishes.
\[ 90 ^\circ \]
Halve
Why: The rotation angle.
\[ 45 ^\circ \]
Figure (svg): Three cards giving the rotation formulas, the angle that removes the cross term, and the special case when the two squared coefficients are equal
\[ \theta=45^\circ \]
Verify: check the discriminant first
Why: The discriminant is 16 minus 36, which is negative 20 — an ellipse. So a 45 degree rotation should produce a standard ellipse equation with no cross term, which is what the substitution gives.
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1261-1262
Prediction
The two squared coefficients are the same.
Predict first
What is the rotation angle?
Correct: Forty-five degrees.
Why: Equal squared coefficients make the cotangent's numerator zero, so twice the angle is ninety degrees and the angle is forty-five. This case is worth recognising on sight, since no computation is needed.
Worked example
The cotangent is nonzero.
\[ \text{Find the rotation angle for } 5x^2-4xy+2y^2-30=0. \]
Read the coefficients
Why: Signs included.
\[ A = 5, B = -4, C = 2 \]
Compute the cotangent
Why: Difference over cross.
\[ \frac{3}{-4} \]
Find twice the angle
Why: By an inverse.
\[ \text{about } 126.9 ^\circ \]
Halve
Why: The rotation angle.
\[ \text{about } 63.4 ^\circ \]
Figure (svg): The solution to Worked example a general angle shown as a ladder of expressions, one row per legal move
\[ \theta\approx 63.4^\circ \]
Verify: check the quadrant
Why: The cotangent is negative, so twice the angle lies in the second quadrant and the angle itself is between 45 and 90 degrees — which 63.4 satisfies. Taking an inverse cotangent's principal value and checking the quadrant is the same care §8.3 required.
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1262-1263
Error analysis
A student finds a rotation angle.
Annotate
On: \( \cot 2\theta=\tfrac{3}{-4} \;\Longrightarrow\; \theta\approx 126.9^\circ \)
The doubling in the formula is easy to lose between reading it and using the result. Writing the intermediate value explicitly as twice the angle keeps it visible.
Faded example
After finding twice the angle to be about 126.9 degrees.
Fill in the blanks
\theta=\frac263.4}\approx___^\circ
Why: The formula gives twice the rotation angle, so the inverse's output must be halved. Skipping this leaves the cross term in place, since the wrong rotation was performed.
Sorting
The angle formula depends on all three.
Sort into buckets
Sort each situation.
Explain it to yourself
Adding a right angle to the solution also removes the cross term.
Discussion prompt
Explain why that is expected.
Hint: What does a further quarter turn do to the axes?
Answer:
Rotating a further right angle swaps the two axes' roles, exchanging which one the conic's major axis lies along.
That produces a different standard form — the same conic described with the roles of the two variables interchanged — but no cross term, since the axes still line up with the conic.
So several angles are valid and they differ in which standard form results. A good explanation notes that the smallest positive angle is conventional rather than uniquely correct.
Section
Section 5
Concept
Particular coefficient combinations produce a point, a line, a pair of lines, or no graph at all. The discriminant names the family but not whether the member is genuine.
The degenerate cases arise when the right side comes out zero or negative after the squares are completed. That is the same phenomenon as a circle equation with a negative radius squared, which has no points at all.
Figure (svg): A contrast between the genuine conics and the degenerate cases a second-degree equation can also describe
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1263-1265
Picture it
The discriminant does not distinguish these.
Figure (svg): A contrast between the genuine conics and the degenerate cases a second-degree equation can also describe
The last row on each side is the practical point. Classifying by discriminant is fast but incomplete, and only the completed squares settle whether the conic is real.
Worked example
The right side comes out zero.
\[ \text{Describe } x^2+4y^2=0. \]
Compute the discriminant
Why: No cross term.
\[ 0 - 16 = -16 \]
Read the family
Why: Negative.
Examine the equation
Why: A sum of squares equals zero.
Conclude
Why: Only the origin.
Figure (svg): A contrast between the genuine conics and the degenerate cases a second-degree equation can also describe
\[ \text{the point }(0,0) \]
Verify: check why no other point works
Why: Both terms are squares and therefore non-negative, so their sum is zero only if each is zero. That forces both coordinates to be zero, leaving exactly one point. The discriminant correctly named the family and said nothing about the degeneracy.
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1263-1264
Prediction
A sum of two squares equals zero.
Predict first
What is the graph?
Correct: A single point.
Why: Squares are non-negative, so a sum of them is zero only when each is zero. That forces both coordinates and leaves exactly one point — a degenerate member of the ellipse family.
Worked example
Two lines rather than two branches.
\[ \text{Describe } x^2-4y^2=0. \]
Compute the discriminant
Why: Positive.
Factor the left side
Why: A difference of squares.
\[ (x - 2 y) (x + 2 y) \]
Set each factor to zero
Why: The zero-product property.
Describe
Why: Two lines through the origin.
Figure (svg): The solution to Worked example a degenerate hyperbola shown as a ladder of expressions, one row per legal move
\[ x=\pm 2y \]
Verify: compare with a genuine hyperbola
Why: These two lines are exactly the asymptotes of the hyperbolas with the same squared coefficients and a nonzero right side. The degenerate case is what remains when the branches collapse onto their own asymptotes.
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1264-1265
Trap
\[ \text{the discriminant is negative, so it is an ellipse} \]
Report the family as the answer
Why: The possibility of degeneracy is not considered.
A single point or an empty graph gets described as an ellipse.
The discriminant names the family, not the member. A negative value covers genuine ellipses, single points and empty graphs alike.
Completing the square and inspecting the right side is what settles it: positive gives a genuine conic, zero or negative gives a degenerate case.
Use the discriminant for speed and the completed squares for certainty. The two answer different questions.
Sorting
Look at the right side after completing squares.
Sort into buckets
Sort each situation.
Faded example
A difference of squares equal to zero.
Fill in the blanks
x^2-4y^2=(x-2y)(x+2y)=0 \;\Longrightarrow\; \textlines___
Why: The zero-product property gives two linear equations, each describing a line through the origin. Those two lines are exactly the asymptotes of the genuine hyperbolas with the same squared coefficients.
Explain it
It classifies quickly but not completely.
Discussion prompt
Explain the limits of the discriminant to a classmate.
Hint: What information does it use?
Answer:
It uses only the three second-degree coefficients — the linear terms and the constant play no part.
But whether a conic is genuine or degenerate depends on those very terms, since they determine what the right side becomes after completing the squares.
So the discriminant is blind to the distinction by construction. A good explanation notes the division of labour: use it to name the family fast, and complete the squares when you need to know whether the conic is real.
Comparison
Fill the blanks from memory. Reading the structure decides the method.
Comparison matrix
| cross term | linear terms | neither | |
|---|---|---|---|
| indicates | the conic is rotated | the conic is translated | centred and aligned |
| the remedy | rotate the axes | complete the square | read it directly |
| order of work | first | second | not needed |
| affects the discriminant | no, it is invariant | no, it is not used | no |
The third row is the practical instruction. Completing the square before rotating does not work, because the cross term prevents the squares from separating.
Pattern
Five steps, and the second often makes the rest unnecessary.
Step 3 is worth stating explicitly. Most questions ask only for the type, and the rotation is a substantial computation to perform unnecessarily.
OpenStax Algebra and Trigonometry 2e, §12.4 Rotation of Axes §12.4
Check
The cross term.
Check your understanding
What does the presence of an xy term indicate?
Answer: A
Why: Substituting a rotation into a standard conic equation produces a cross term, and removing it requires rotating the axes back. Translation produces linear terms instead.
Check
The discriminant.
Check your understanding
A discriminant of zero identifies which conic?
Answer: A
Why: Zero is the boundary between the negative case, an ellipse, and the positive case, a hyperbola — and it corresponds to the second-degree terms forming a perfect square, which is the parabola's algebraic signature.
Check
Degenerate cases.
Check your understanding
What does the discriminant fail to tell you?
Answer: A
Why: It uses only the second-degree coefficients, and degeneracy depends on the linear terms and the constant. A negative value covers genuine ellipses, single points and empty graphs alike.
Real world
Structural engineers use invariants exactly as this section uses the discriminant.
Discussion prompt
Stress at a point in a material is described by coefficients that change with the chosen axes. Why do engineers compute invariants?
Hint: What should not depend on how you set up the coordinates?
Answer:
The individual coefficients depend on which directions the axes were chosen along, which is an arbitrary decision rather than a fact about the material.
So engineers compute combinations that are unchanged by rotating the axes — invariants — because only those describe the physical state rather than the bookkeeping.
Whether a material yields depends on such an invariant, not on any single coefficient. The discriminant is the same idea in a simpler setting: a combination that survives the change of frame and therefore says something about the object rather than the description.
Commit first
State your confidence along with your answer.
Predict first
Why is the discriminant useful for classifying a rotated conic?
Correct: It is unchanged by rotation, so no rotation is needed.
Why: The three second-degree coefficients all change under rotation, but this particular combination of them does not. That invariance is what lets the tilted equation be classified directly, saving the whole rotation computation.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate is about to rotate a conic just to find out what type it is. What do you tell them?
Hint: Is there a shortcut?
Answer:
The discriminant answers that in one computation, without any rotation at all.
It works because the quantity is unchanged by rotating the axes, so computing it on the tilted equation gives the same answer as computing it after rotating.
Rotation is only needed when the conic's actual features are wanted — its centre, axes, foci. A good explanation adds the caveat: the discriminant names the family but not whether the conic is genuine, so a degenerate case can still be lurking.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second is the section's most useful single tool, since it answers the commonest question with the least work. The third is where the arithmetic is, particularly the halving of the angle.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the general second-degree equation and label what each kind of term indicates about position and orientation. Beside it, write the discriminant with its three cases and note that it is unchanged by rotation. Underneath, work one rotation from angle to standard form, and list the degenerate cases the discriminant cannot detect.
If your discriminant note says explicitly that it names the family but not the member, the section's one real limitation is on the page rather than assumed away.
Recap
Five things, and the second saves the most work.
| if you remember one thing | it should be this |
|---|---|
| about the cross term | it means rotated, and rotation comes before completing squares |
| about the discriminant | negative ellipse, zero parabola, positive hyperbola |
| about the angle | the formula gives twice it, so halve |
| about degeneracy | the discriminant names the family, not the member |
Section 10.5 closes the chapter by describing all three conics with a single polar equation, in which the eccentricity alone distinguishes them.
OpenStax, Precalculus, §10.4 Rotation of Axes §10.4, pp. 1248-1265 — everything on these slides traces back here
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