The standard forms of translated conic sections, graphing a translated circle and hyperbola, writing equations of translated parabolas and ellipses from their foci and vertices, identifying lines of symmetry, and classifying a general second-degree equation using the discriminant.
Subject: Algebra 2 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 2 · Chapter 9 — Quadratic Relations and Conic Sections
Translate and Classify Conic Sections
Objectives
Five outcomes. The four conics moved off the origin, and one test that names any of them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 650-655 — the lesson these objectives are drawn from
Warm-up
Chapter 4 shifted a parabola by writing x minus h in place of x, and Chapters 6 and 7 did the same for radicals and logarithms.
Discussion prompt
The graph of y equals x squared has vertex at the origin. Where is the vertex of y equals the quantity x minus 2, squared, plus 5? And what would you expect x minus 2 to do inside a circle's equation?
Hint: Subtracting inside shifts the other way.
Answer:
\[ y = (x-2)^2+5: \; \text{vertex } (2, 5) \]
\[ (x-2)^2+(y+3)^2 = 9: \; \text{centre } (2,-3) \]
Exactly the same grammar. Every conic of this chapter translates by the same substitution, so nothing new has to be learned — only applied to four more families at once.
Concept
Replacing x by x minus h and y by y minus k moves any conic so that its centre, or a parabola's vertex, sits at the point h comma k. Every other feature keeps its distance from that point.
conic sections — The four curves formed when a plane cuts a double cone: parabolas, circles, ellipses and hyperbolas. Any of them can be described by a general second-degree equation in x and y.
\[ (x-h)^2+(y-k)^2 = r^2 \]
A general second-degree equation may be any of the four, and the discriminant B squared minus 4AC identifies which without any rearranging at all.
Figure (svg): Two columns comparing an origin-centred conic with a translated one
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 650-653
Section
Section 1
Concept
For every conic, replacing x by x minus h and y by y minus k moves the whole curve so its centre or vertex is at h comma k. The letters a, b, p and r keep their meanings exactly.
\[ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \]
The sign inside each bracket is opposite to the coordinate it produces, which is the same reversal met with every translated family since Chapter 4.
Figure (svg): The four conic standard forms rewritten with a centre or vertex at h comma k
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 650-650 — Standard Form of Equations of Translated Conics
Picture it
The standard forms with a centre or vertex at h comma k.
Figure (svg): The four conic standard forms rewritten with a centre or vertex at h comma k
Only the brackets changed. Each family's own structure — a sum, a difference, a single square — is exactly as before.
Worked example
Applying the substitution to the guided practice equations.
\[ \text{Find the centre or vertex of } (x+1)^2+(y-3)^2=4, \; (x-2)^2=8(y+3), \; (x+3)^2-\tfrac{(y-4)^2}{4}=1, \; \tfrac{(x-2)^2}{16}+\tfrac{(y-1)^2}{9}=1. \]
First: a circle
Why: X plus 1 gives h equal to negative 1; y minus 3 gives k equal to 3.
\[ \text{centre } (-1, 3), r = 2 \]
Second: a parabola
Why: X minus 2 gives h equal to 2; y plus 3 gives k equal to negative 3.
\[ \text{vertex } (2, -3) \]
Third: a hyperbola
Why: X plus 3 gives negative 3; y minus 4 gives 4.
\[ \text{centre } (-3, 4) \]
Fourth: an ellipse
Why: X minus 2 gives 2; y minus 1 gives 1.
\[ \text{centre } (2, 1) \]
Figure (svg): The four conic standard forms rewritten with a centre or vertex at h comma k
\[ (-1,3), \; (2,-3), \; (-3,4), \; (2,1) \]
Verify: check each sign reversal
Why: Every plus inside a bracket produced a negative coordinate and every minus produced a positive one. Substituting the centre back should make each bracket zero: at negative 1 comma 3 the first equation's brackets are 0 and 0, giving 0 equal to 4 — false, which is right, since the centre is not ON the circle.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 651-651
Fill the middle
Example 1.
Fill in the blanks
y+3 = 0 \;\Longrightarrow\; k = -3
Why: Setting y plus 3 equal to zero gives k equal to negative 3, so the centre sits three units below the horizontal axis. Reading the 3 directly would put it three units above.
Worked example
Reading the structure rather than the numbers.
\[ \text{Name the family of each of the four equations above, and say why.} \]
Two squares added, equal to a constant
Why: Both coefficients are 1, so all radii match.
One square, equal to a linear term
Why: Only x is squared.
Two squares subtracted, equal to 1
Why: The minus sign is decisive.
Two squares added, equal to 1, different denominators
Why: Unequal denominators rule out a circle.
Figure (svg): The solution to Worked example identify each family shown as a ladder of expressions, one row per algebraic move
\[ \text{circle}, \; \text{parabola}, \; \text{hyperbola}, \; \text{ellipse} \]
Verify: state the distinguishing feature of each
Why: One square means a parabola. Two squares added means a circle if the denominators match and an ellipse otherwise. Two squares subtracted means a hyperbola. That decision tree handles every equation already in standard form, and the discriminant of the last idea handles the ones that are not.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 650-650
Trap
\[ (x-2)^2+(y+3)^2 = 9 \]
Read the centre as negative 2 comma 3
Why: The signs are copied straight from the brackets.
\[ \text{centre } (-2, 3) \quad \text{(wrong)} \]
Substituting negative 2 comma 3 gives 16 plus 36, or 52, not 0 — so that point is not the centre and is not even on the circle.
\[ x-h = x-2 \;\Longrightarrow\; h = 2; \quad y-k = y+3 \;\Longrightarrow\; k = -3 \]
Solve each bracket for zero
Why: The centre is where both brackets vanish.
\[ \text{centre } (2, -3) \]
Setting each bracket equal to zero is the reliable move, and it is the same one that located every asymptote and vertex in earlier chapters.
Matching
Set each bracket to zero.
Match the pairs
Why: The first and third share the same point but call it by different names: a circle has a centre and a parabola has a vertex. The point plays the same role in both, as the anchor everything else is measured from.
Sorting
Count the squares and read the sign.
Sort into buckets
Sort each equation by its family.
Counting squares first, then reading the sign, settles every equation already in standard form in a couple of seconds.
Prediction
Commit before reasoning.
Predict first
How does the hyperbola with centre at negative 1 comma 3 differ in shape from the one with the same a and b at the origin?
Correct: Not at all; only its position has changed.
\[ \text{asymptotes: } y-3 = \pm\tfrac{2}{3}(x+1) \]
Why: A translation slides every point by the same amount, so distances, slopes and proportions are all preserved. The asymptotes have the same slopes as before — two thirds here — but pass through the new centre rather than the origin. This is why translated problems are solved by finding the centre and then treating everything else exactly as in the previous lessons.
Section
Section 2
Concept
Read the centre or vertex from the brackets, plot it, and then place every other feature at its usual distance from that point rather than from the origin.
\[ \text{centre } (-1,3), \; a = 2, \; b = 3 \]
For a hyperbola that means drawing the central rectangle around the new centre; for an ellipse, stepping a and b out from it; for a circle, stepping r in four directions.
Figure (svg): A translated hyperbola with its centre, vertices, foci and central rectangle
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 650-651 — Graph the equation of a translated conic
Picture it
Example 2, with its rectangle and asymptotes around the new centre.
Figure (svg): A translated hyperbola with its centre, vertices, foci and central rectangle
The vertices sit 2 above and below the centre and the foci about 3.6 above and below. Everything is measured from negative 1 comma 3.
Worked example
Example 1.
\[ \text{Graph } (x-2)^2+(y+3)^2 = 9. \]
Read the centre
Why: Set each bracket to zero.
\[ (2, -3) \]
Read the radius
Why: The square root of 9.
\[ r = 3 \]
Step 3 units in four directions
Why: Right, left, up and down from the centre.
\[ (5, -3), (-1, -3), (2, 0), (2, -6) \]
Draw the circle
Why: Through those four points.
\[ a\text{ circle of radius } 3 \]
Figure (svg): A circle translated away from the origin, with its centre and four surrounding points
\[ \text{centre } (2,-3), \; r = 3 \]
Verify: test one of the four points
Why: At 2 comma 0 the equation gives 0 plus 9, which is 9 — correct. Note that the four points are found by adding and subtracting r from each coordinate of the centre in turn, never from the origin.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 650-650
Fill the middle
Example 2.
Fill in the blanks
\text1 (-1,3), \; a = 2 \;\Longrightarrow\; \text___ (-1, 5) \text___ (-1, ___)
Why: Three minus 2 is 1, so the lower vertex is at negative 1 comma 1. Both vertices are found by adding and subtracting a from the centre's y-coordinate.
Worked example
Example 2.
\[ \text{Graph } \frac{(y-3)^2}{4}-\frac{(x+1)^2}{9}=1. \]
Read the centre and orientation
Why: Brackets give negative 1 comma 3; the y term is positive.
\[ \text{centre } (-1, 3),\text{ vertical} \]
Read a and b
Why: The square roots of 4 and 9.
\[ a = 2, b = 3 \]
Place the vertices and foci
Why: Two above and below; c squared is 13, so about 3.6.
\[ (-1, 5), (-1, 1); (-1, 6.6), (-1, -0.6) \]
Draw the rectangle and branches
Why: Four tall and 6 wide, centred at the centre.
Figure (svg): A translated hyperbola with its centre, vertices, foci and central rectangle
\[ (-1,3); \; (-1,5),(-1,1); \; (-1,\,3\pm\sqrt{13}) \]
Verify: check a vertex in the equation
Why: At negative 1 comma 5 the equation gives 4 over 4 minus 0, which is 1 — so that vertex is on the curve. And the foci lie farther from the centre than the vertices, as a hyperbola requires, since the square root of 13 exceeds 2.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 651-651
Error analysis
A student graphs a hyperbola centred at negative 1 comma 3 with a equal to 2.
Annotate
On: \( \text{vertices at } (0, 2) \text{ and } (0, -2) \)
After a translation, every distance is measured from the new centre. Plotting the centre first and working outward from it makes the mistake almost impossible.
Ranking
Graphing any translated conic.
Put in order
Why: Steps one to three could be done in any order, but step four cannot begin until the centre is known. Plotting the centre before anything else is what keeps every subsequent distance measured from the right place.
Comparison
Fill the blanks. The same hyperbola, moved.
Comparison matrix
| Feature | Centred at the origin | Centred at (-1, 3) |
|---|---|---|
| Centre | (0, 0) | (-1, 3) |
| Vertices | (0, +-2) | (-1, 5) and (-1, 1) |
| Asymptote slopes | +-2/3, unchanged | +-2/3 |
| Asymptote equations | y = +-(2/3)x | y - 3 = +-(2/3)(x + 1) |
The slopes survive the translation untouched, but the lines they describe move with the centre — so the equations pick up the same brackets as the conic itself.
Sorting
A translation slides everything.
Sort into buckets
Sort each feature by whether translating changes it.
Positions move and measurements do not, which is exactly what makes translated problems no harder than untranslated ones once the centre is found.
Section
Section 3
Concept
The centre of an ellipse or hyperbola is the midpoint of its two foci, or of its two vertices. A parabola's vertex is given directly, and p is the signed distance from it to the focus.
\[ (h,k) = \left(\frac{1+7}{2}, \frac{2+2}{2}\right) = (4,2) \]
Because the foci and co-vertices lie on lines through the centre, the distances b and c are single subtractions rather than applications of the distance formula.
Figure (svg): A translated parabola and a translated ellipse built from their given points
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 651-652 — Write an equation of a translated conic
Picture it
Examples 3 and 4, built from their given points.
Figure (svg): A translated parabola and a translated ellipse built from their given points
The parabola's focus is left of its vertex, so it opens left with p negative. The ellipse's centre is the midpoint of its foci, at 4 comma 2.
Worked example
Example 3.
\[ \text{Write the parabola with vertex } (-2,3) \text{ and focus } (-4,3). \]
Decide the form
Why: The focus is left of the vertex, so it opens left.
\[ (y - k) ^{2} = 4 p(x - h) \]
Read h and k
Why: The vertex gives both directly.
\[ h = -2, k = 3 \]
Find the size of p
Why: The distance from vertex to focus along the line y equal to 3.
\[ | p | = 2 \]
Fix the sign of p
Why: Opening left means p is negative.
\[ p = -2,\text{ so } 4 p = -8 \]
Figure (svg): A translated parabola and a translated ellipse built from their given points
\[ (y-3)^2 = -8(x+2) \]
Verify: check the vertex and the direction
Why: At negative 2 comma 3 both sides are zero, so the vertex is on the curve. And for the left side to be a square, x plus 2 must be at most zero, so x is at most negative 2 — the curve really does lie to the left of its vertex, matching the focus's position.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 651-651
Fill the middle
Example 4.
Fill in the blanks
(h,k) = \left(\frac4___, \frac______\right) = (___, 2)
Why: The midpoint of 1 and 7 is 4, so the centre is 4 comma 2. It is also the midpoint of the two co-vertices, which gives the same point and confirms the reading.
Worked example
Example 4 and Guided Practice 5 and 6.
\[ \text{Write the ellipse with foci } (1,2),(7,2) \text{ and co-vertices } (4,0),(4,4). \]
Find the centre
Why: The midpoint of the two foci.
\[ (4, 2) \]
Read the orientation
Why: The foci lie on a horizontal line.
Find b and c by subtraction
Why: From the centre to a co-vertex, and to a focus.
\[ b = 2, c = 3 \]
Find a squared
Why: Four plus 9, since for an ellipse a squared is b squared plus c squared.
\[ a ^{2} = 13 \]
Figure (svg): A translated parabola and a translated ellipse built from their given points
\[ \frac{(x-4)^2}{13}+\frac{(y-2)^2}{4}=1 \]
Verify: check a co-vertex
Why: At 4 comma 4 the equation gives 0 plus 4 over 4, which is 1 — so that co-vertex is on the ellipse. And a squared, 13, exceeds b squared, 4, as it must when the major axis is horizontal. The co-vertices and foci lay on vertical and horizontal lines through the centre, so no distance formula was needed anywhere.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 652-652
Trap
\[ \text{vertices } (-7,3) \text{ and } (-1,3) \]
Take negative 7 comma 3 as the centre
Why: The first named point is used as the anchor.
\[ \text{centre } (-7,3) \quad \text{(wrong)} \]
Then the other vertex would be 6 units away and there would be nothing 6 units the other side, so the curve would not be symmetric.
\[ (h,k) = \left(\tfrac{-7+(-1)}{2}, 3\right) = (-4, 3) \]
The centre is the midpoint of the two vertices
Why: It is also the midpoint of the two foci, which is a useful cross-check.
\[ a = 3, \; c = 5 \;\Longrightarrow\; b^2 = 16 \]
Computing the centre both ways, from the vertices and from the foci, should give the same point — and if it does not, one of the given points has been misread.
Matching
Centre first, then the constants.
Match the pairs
Why: The two parabolas took p directly from the vertex-to-focus distance with a sign for direction. The ellipse and hyperbola both needed a midpoint first, and then differed only in whether the relation for the third constant added or subtracted.
Sorting
The family decides the sign.
Sort into buckets
Sort each situation.
The two relations look different but are the same equation rearranged, with the largest quantity alone on one side in each case.
Prediction
Commit before reasoning.
Predict first
Example 4 finds b and c by simple subtraction rather than with the distance formula. Why is that legitimate?
Correct: Because the co-vertices and foci lie on vertical and horizontal lines through the centre.
\[ \text{from } (4,2) \text{ to } (1,2): \; |1-4| = 3 \]
Why: Along a horizontal line only the x-coordinates differ, so the distance is a single subtraction — and the distance formula would return the same value after squaring a zero and taking a root. The axes of a conic are always horizontal or vertical in this chapter, which is why the shortcut always applies. The book prints exactly this observation beside Example 4.
Section
Section 4
Concept
A circle is symmetric about every line through its centre. An ellipse and a hyperbola are symmetric about the two lines through the centre parallel to the axes. A parabola has one line of symmetry, through its vertex and focus.
\[ x = h \quad \text{and} \quad y = k \]
The lines are always x equal to h and y equal to k, so reading the centre gives the symmetry lines immediately with no further work.
Figure (svg): A translated parabola and a translated ellipse built from their given points
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 652-652 — Identify symmetries of conic sections
Picture it
The conics of Examples 3 and 4 with their symmetry lines implied by their centres.
Figure (svg): A translated parabola and a translated ellipse built from their given points
The parabola is symmetric about y equal to 3 only. The ellipse is symmetric about both x equal to 4 and y equal to 2.
Worked example
Example 5, for all four earlier conics.
\[ \text{Name the lines of symmetry for the circle at } (2,-3), \text{ the hyperbola at } (-1,3), \text{ the parabola with vertex } (-2,3), \text{ and the ellipse at } (4,2). \]
The circle
Why: Every line through the centre works.
The hyperbola
Why: Both lines through the centre parallel to the axes.
\[ x = -1\text{ and } y = 3 \]
The parabola
Why: Only the line through vertex and focus.
\[ y = 3 \]
The ellipse
Why: Both lines through the centre.
\[ x = 4\text{ and } y = 2 \]
Figure (svg): The solution to Worked example name the symmetry lines shown as a ladder of expressions, one row per algebraic move
\[ \text{all}; \; x=-1, y=3; \; y=3; \; x=4, y=2 \]
Verify: explain why the parabola has only one
Why: A parabola opens in one direction only, so reflecting across the line perpendicular to its axis would send the branch the wrong way. The other three families are symmetric front to back as well as side to side, which is why they get two lines — and the circle, being the same in every direction, gets all of them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 652-652
Sorting
Count what the shape allows.
Sort into buckets
Sort each conic.
The count is a quick way to check a classification: an answer with two symmetry lines cannot be a parabola.
Worked example
Guided Practice 7 to 9.
\[ \text{Name the symmetry lines of } \tfrac{(x-5)^2}{64}+\tfrac{y^2}{16}=1, \; (x+5)^2=8(y-2), \; \tfrac{(x-1)^2}{49}-\tfrac{(y-2)^2}{121}=1. \]
First: an ellipse
Why: Centre at 5 comma 0.
\[ x = 5\text{ and } y = 0 \]
Second: a parabola
Why: Vertex at negative 5 comma 2, opening up.
\[ x = -5\text{ only} \]
Third: a hyperbola
Why: Centre at 1 comma 2.
\[ x = 1\text{ and } y = 2 \]
Note the pattern
Why: Two lines for centred conics, one for a parabola.
Figure (svg): The solution to Worked example three more from equations shown as a ladder of expressions, one row per algebraic move
\[ x=5, y=0; \; x=-5; \; x=1, y=2 \]
Verify: check the parabola's direction
Why: The second has x squared, so its axis of symmetry is vertical — the line x equal to negative 5 through its vertex. Had y been squared instead, the single symmetry line would have been horizontal. Which variable carries the square decides which line it is.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 653-653
Error analysis
A student names the symmetry lines of a parabola with vertex at negative 5 comma 2.
Annotate
On: \( x = -5 \text{ and } y = 2 \)
Only the closed and two-branched conics have a second line. A parabola is unbounded in one direction only, which allows just one reflection.
Fill the middle
Guided Practice 7.
Fill in the blanks
\tfrac5___+\tfrac______=1: \; x = ___ \text___ y = 0
Why: The centre is at 5 comma 0, so the two symmetry lines are x equal to 5 and y equal to 0. Reading the centre gives both at once.
Matching
Read the centre or vertex.
Match the pairs
Why: The parabola's single line is horizontal because its vertex and focus share a y-coordinate, so it opens sideways. A parabola's one symmetry line always runs through both.
Prediction
Commit before reasoning.
Predict first
Why is every line through a circle's centre a line of symmetry?
Correct: Because every point is the same distance from the centre, so reflection across any such line preserves the distance.
\[ a = b \;\Longrightarrow\; \text{no preferred direction} \]
Why: Reflecting across a line through the centre keeps every point's distance from the centre unchanged, so a point of the circle maps to another point of the circle. An ellipse fails this for most lines because its distance from the centre varies with direction, leaving only the two axes. The circle's total symmetry is another way of saying that a equals b — the same fact that made c equal to zero in Lesson 9.4.
Section
Section 5
Concept
Any conic can be written as A x squared plus B xy plus C y squared plus D x plus E y plus F equal to zero. The discriminant B squared minus 4AC identifies the family: negative for a circle or ellipse, zero for a parabola, positive for a hyperbola.
\[ B^2-4AC \]
A circle is the case where the discriminant is negative, B is zero and A equals C. Any other negative discriminant gives an ellipse.
Figure (svg): The discriminant test for classifying a second-degree equation, with one worked case
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 653-653 — Classifying Conics Using Their Equations
Picture it
Example 6: classified, then completed and graphed.
Figure (svg): The discriminant test for classifying a second-degree equation, with one worked case
The discriminant is negative 16 with A not equal to C, so an ellipse. Completing the square in x gives the standard form and the graph.
Worked example
Example 6.
\[ \text{Classify and graph } 4x^2+y^2-8x-8=0. \]
Read A, B and C
Why: Four, zero and 1.
\[ A = 4, B = 0, C = 1 \]
Compute the discriminant
Why: Zero minus 4 times 4 times 1.
\[ -16 \]
Classify
Why: Negative, and A is not equal to C.
Complete the square in x
Why: Factor 4 out, add 1 inside and 4 to the right.
\[ (x - 1) ^{2} / 3 + y ^{2} / 12 = 1 \]
Figure (svg): The discriminant test for classifying a second-degree equation, with one worked case
\[ \frac{(x-1)^2}{3}+\frac{y^2}{12}=1 \]
Verify: check the completing of the square
Why: Expanding 4 times x minus 1 squared gives 4x squared minus 8x plus 4, and the original had 4x squared minus 8x — so 4 was added on the left and 4 must be added on the right too, taking 8 to 12. Adding 1 inside a bracket multiplied by 4 adds 4 overall, which is the step most often mishandled.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 653-653
Fill the middle
Example 6.
Fill in the blanks
B^2-4AC = 0^2-4(4)(1) = -16
Why: The discriminant is negative 16, and since A is 4 while C is 1 the conic is an ellipse rather than a circle. Both conditions have to be checked for the circle case.
Worked example
Applying the discriminant test without rearranging.
\[ \text{Classify } x^2+y^2-4x=0, \; 4x^2+y^2-8x-8=0, \; x^2-4y+8=0, \; x^2-y^2-6x=0. \]
First: A is 1, B is 0, C is 1
Why: Discriminant negative 4, and A equals C.
Second: A is 4, C is 1
Why: Discriminant negative 16, and A is not C.
Third: C is 0
Why: Discriminant is zero.
Fourth: A is 1, C is negative 1
Why: Zero minus 4 times 1 times negative 1 is 4.
Figure (svg): The solution to Worked example classify four equations shown as a ladder of expressions, one row per algebraic move
\[ \text{circle}, \; \text{ellipse}, \; \text{parabola}, \; \text{hyperbola} \]
Verify: notice what the test does not need
Why: None of the four had to be rearranged, completed or graphed — three coefficients were enough. The test also explains the families structurally: a parabola has only one squared term so one of A and C is zero, and a hyperbola has squared terms of opposite signs so their product is negative.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 653-653
Trap
\[ 4(x^2-2x+\;?\;)+y^2 = 8+\;? \]
Add 1 inside the bracket and 1 to the right
Why: The same number is added to both sides.
\[ 4(x-1)^2+y^2 = 9 \quad \text{(wrong)} \]
The 1 inside is multiplied by the 4 outside, so 4 was added on the left and only 1 on the right.
\[ 4(x^2-2x+1)+y^2 = 8+4(1) \]
Multiply the added amount by the factor outside
Why: Whatever the bracket is multiplied by, the addition is multiplied by too.
\[ 4(x-1)^2+y^2 = 12 \]
This is the same care Lesson 4.7 required when completing the square with a leading coefficient. The factor outside always scales what goes in.
Sorting
Compute B squared minus 4AC.
Sort into buckets
Sort each general equation.
Three coefficients settle all five, with no rearranging and no graphing — which is why the test is worth applying before any algebra.
Two truths and a lie
All three are about classifying conics.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The discriminant uses only A, B and C, and the centre depends on D, E and F as well — which is why completing the square is still needed after classifying. The test names the family and nothing more, but naming the family first tells you which standard form you are aiming at.
Prediction
Commit before reasoning.
Predict first
With B equal to zero, when is B squared minus 4AC positive?
Correct: When A and C have opposite signs, so their product is negative.
\[ B = 0: \; B^2-4AC = -4AC > 0 \;\Longleftrightarrow\; AC < 0 \]
Why: With B equal to zero the discriminant is negative 4 times A times C, which is positive exactly when the product AC is negative — that is, when the two squared terms have opposite signs. And opposite signs is precisely what makes a difference of squares, which is a hyperbola. So the test is not an arbitrary formula but a compact statement of the structural difference between the families.
Comparison
Fill the blanks. Structure, symmetry and discriminant all agree.
Comparison matrix
| Family | Structure | Lines of symmetry |
|---|---|---|
| Circle | two squares added, equal coefficients | infinitely many |
| Parabola | only one variable squared | exactly one |
| Ellipse | two squares added, unequal denominators | exactly two |
| Hyperbola | two squares subtracted | exactly two |
Three independent readings — the structure, the symmetry count and the discriminant — all give the same answer, so any two can check the third.
Pattern
One routine to graph, one to write, one to classify.
When completing the square inside a bracket with a coefficient outside, the amount added to the other side is that coefficient times the amount added inside.
OpenStax Algebra and Trigonometry 2e, §12.4 Rotation of Axes §12.4
Check
Set each bracket to zero.
Check your understanding
What is the centre of (x - 2)^2 + (y + 3)^2 = 9?
Answer: A
Why: Solving x - 2 = 0 and y + 3 = 0 gives 2 and -3.
Check
Find the centre before anything else.
Check your understanding
Write the ellipse with foci (1,2) and (7,2) and co-vertices (4,0) and (4,4).
Answer: A
Why: The centre is (4,2), with b = 2 and c = 3, so a squared is 4 + 9.
Check
Three coefficients are enough.
Check your understanding
Classify 4x^2 + y^2 - 8x - 8 = 0.
Answer: A
Why: The discriminant is -16, which is negative, and A is 4 while C is 1.
Real world
A whispering gallery is a room with an elliptical ceiling in which a whisper at one focus is heard clearly at the other. One such gallery is 96 feet long and 46 feet high at the centre, with the ceiling forming half an ellipse.
Discussion prompt
Taking the centre of the floor as the origin, write the equation of the ceiling and find where the two listening spots are.
Hint: The semi-axes are half the length and the full height.
Answer:
\[ a = \tfrac{96}{2} = 48, \quad b = 46 \]
\[ \frac{x^2}{2304}+\frac{y^2}{2116} = 1 \]
\[ c^2 = 2304-2116 = 188 \;\Longrightarrow\; c \approx 13.7 \]
The two spots are about 13.7 feet either side of the centre, on the floor. Standing at one and whispering, every sound wave that leaves you reflects off the ceiling and arrives at the other spot — and because the two path lengths from focus to ceiling to focus always add to the same 2a, they arrive together rather than smeared out.
That is the constant-sum definition doing real acoustic work: it is not merely that the sound is directed to the second focus, but that every route takes the same time. The Statuary Hall in the United States Capitol works this way, and the story goes that one representative used to nap at the right spot to eavesdrop on the opposition. Real galleries are translated versions of this equation, with the centre wherever the architect put it, which is exactly what this lesson supplies.
Commit first
Answer, then rate your confidence honestly.
Predict first
In x minus 2, all squared, plus y plus 3, all squared, equal to 9, is the centre at negative 2 comma 3?
Correct: No — solve each bracket for zero, giving 2 and negative 3.
\[ y+3 = y-(-3) \;\Longrightarrow\; k = -3 \]
Why: The standard form is x minus h and y minus k, so a written plus 3 means k is negative 3. Substituting settles it: at 2 comma negative 3 both brackets vanish, and stepping 3 units in any direction from there lands on the circle, while negative 2 comma 3 gives 16 plus 36, or 52, nowhere near 9. The same reversal has appeared in every translated family since Chapter 4, and solving each bracket for zero avoids it every time.
Explain it
They can graph a circle centred at the origin and are alarmed by brackets.
Discussion prompt
In four sentences or fewer, explain how to graph a circle whose equation has brackets in it.
Hint: What makes each bracket zero?
Answer:
Find the number that makes each bracket equal zero — that pair is the centre. For x minus 2 that number is 2, and for y plus 3 it is negative 3, so the centre is at 2 comma negative 3.
Then take the square root of the number on the right to get the radius, and step that far up, down, left and right from the centre. Draw the circle through those four points; it is the same circle as before, just moved.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For signs, solve each bracket for zero rather than reading it. For measuring, plot the centre before anything else and work outward from it. For completing the square, multiply whatever you add inside by the coefficient outside before adding it to the other side. For the discriminant, remember that a circle needs all three conditions and everything else needs only the sign.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a translation page. Top: write all four translated standard forms in a column, and beside each note what h comma k is called for that family and how many lines of symmetry it has. Middle left: graph Example 1 with the centre plotted first and the four points stepped out from it. Middle right: graph Example 2 in full, drawing the rectangle around the new centre and the asymptotes through its corners, and write both asymptote equations. Bottom left: work Examples 3 and 4, marking on each sketch how you found the centre or vertex. Bottom right: write the discriminant table, classify four equations of your own with it, and complete the square on one of them all the way to standard form.
If any feature in your graphs is measured from the origin rather than from the centre, redo that one: after a translation, the origin has no special role at all.
Recap
Five things, and the four conics are now movable.
| If you see | Then |
|---|---|
| A bracket like x minus h | Solve it for zero to get the coordinate |
| A translated conic | Plot the centre and measure everything from it |
| Two foci or two vertices | Their midpoint is the centre |
| A parabola | One line of symmetry; other conics have two or more |
| A general second-degree equation | Compute B squared minus 4AC first |
| A coefficient outside a completed square | Multiply the added amount by it before balancing |
Lesson 9.7 solves systems in which two conics meet, using substitution and elimination on second-degree equations.
McDougal Littell Algebra 2 (Texas Edition), Ch. 9 Quadratic Relations and Conic Sections — Lesson 9.6 Translate and Classify Conic Sections §9.6, pp. 650-655 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 2 — $55/session, free consultation.