Identifying and translating conics
An equation in x and y with squared terms draws some curve in the plane. Which curve is it, and where does it sit, before anything is graphed?
Where you are 6 in Conic sections
When you reach for it
Each of the four Conic sections has its own standard form, taught in its own lesson: the circle, the ellipse, the hyperbola and the parabola. A standard form tells you at a glance what the curve is and where it sits. But an equation rarely arrives that way. Expand any standard form, move everything to one side, and you get an equation of the shape
with A and C not both zero. This is the general second degree equation (with no xy term; see Boundaries). It hides the type and the position of the curve. This lesson undoes the hiding in two moves. Day 1: read the type from A and C alone. Day 2: complete the square to put the equation back in standard form and read its center or vertex.
The procedure
Day 1: the squared terms name the curve
Tidy the equation first: expand every product, move every term to the left side, and combine like terms. Only then read the coefficients A of x^2 and C of y^2.
To see why these two numbers decide everything, expand each standard form and look at its squared terms. Expanding (x - h)^2 gives x^2 - 2hx + h^2: the square produces one x^2, and the shift h only feeds the x term and the constant. So the shifts never touch A or C.
- A circle (x - h)^2 + (y - k)^2 = r^2 expands with x^2 + y^2: A = C.
- An ellipse \displaystyle \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1, multiplied by a^2b^2, starts b^2x^2 + a^2y^2: A and C have the same sign, so AC > 0.
- A hyperbola \displaystyle \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 starts b^2x^2 - a^2y^2, and the form that opens up and down starts -b^2x^2 + a^2y^2: opposite signs, so AC < 0.
- A parabola (x - h)^2 = 4p(y - k) or (y - k)^2 = 4p(x - h) squares only one variable: one of A, C is zero, so AC = 0.
| The squared terms | The curve, if the equation draws one |
|---|---|
| A = C | circle |
| AC > 0 and A \ne C | ellipse |
| AC < 0 | hyperbola |
| AC = 0 (exactly one of A, C is zero) | parabola |
A circle is the ellipse with A = C, which is why it gets its own row. The table reads correctly in the other direction too: Day 2 shows that any equation of this shape can be rewritten as one of the standard forms with the same A and C, unless it collapses to a point, to lines, or to nothing at all. Those collapsed cases are in Boundaries; the phrase “if the equation draws one” in the table is there for them. The numbers D, E, F never change the type. They move the curve and set its size.
Day 2: completing the square moves the curve home
Replacing x by x - h and y by y - k slides a graph h to the right and k up, as in shifting a graph up, down, left and right. So every standard form with center (h, k) is the curve centered at the origin, slid. To find h and k from a general equation, complete the square in each variable that is squared. Factor out the leading coefficient first, as in completing the square when the leading coefficient is not 1:
The last step is allowed because \displaystyle A\left(x + \frac{D}{2A}\right)^2 expands to \displaystyle Ax^2 + Dx + \frac{D^2}{4A}, so subtracting \displaystyle \frac{D^2}{4A} gives back exactly what we started with. The same holds for y. When both A and C are nonzero, the equation becomes
where \displaystyle G = \frac{D^2}{4A} + \frac{E^2}{4C} - F. The coefficients outside the squares are still A and C, which is the promise Day 1 relied on. Dividing by G (when G \ne 0) gives the standard form of an ellipse, circle or hyperbola, with (h, k) as its center.
When only x is squared, complete the square in x alone and collect the rest: A(x - h)^2 = -E(y - k), a parabola with vertex (h, k). Solve for k by factoring -E out of the right side. When only y is squared, the roles swap.
In practice nobody memorizes \displaystyle h = -\frac{D}{2A}. Completing the square in the equation itself, adding the same amount to both sides, is shorter and shows every step.
Worked examples
Day 1: four equations, four curves
Name the conic each equation draws.
(a) 2x^2 + 2y^2 + 12x - 4y - 5 = 0, (b) 5x^2 + y^2 - 10x + 4y - 11 = 0, (c) y^2 - 6y - 8x + 1 = 0, (d) 4y^2 - x^2 + 2x + 16y + 11 = 0.
SolutionEach equation is already tidy, so read A and C.
(a) A = C = 2: a circle. Dividing by 2 would give x^2 + y^2, the circle’s signature.
(b) A = 5, C = 1: both positive but unequal, so AC = 5 > 0 and the curve is an ellipse, stretched more in one direction than the other.
(c) There is no x^2 term, so A = 0 and C = 1: a parabola. The squared variable is y, so it opens sideways.
(d) The x^2 term is -x^2, so A = -1 and C = 4: AC = -4 < 0, a hyperbola.
Completing the square confirms each one is a genuine curve: (a) becomes \displaystyle (x + 3)^2 + (y - 1)^2 = \frac{25}{2}, (b) becomes 5(x - 1)^2 + (y + 2)^2 = 20, (c) becomes (y - 3)^2 = 8(x + 1), and (d) becomes 4(y + 2)^2 - (x - 1)^2 = 4.
Day 1: tidy before you read
Name the graph of each equation.
(a) x(x + 4) = 3y^2 - 2y + 7, (b) 9 - (x - 1)^2 = (y + 2)^2, (c) y^2 + 4x = (y - 3)^2.
SolutionThe table only applies once every term is on one side and like terms are combined.
(a) Expand the left side and move the right side over: x^2 + 4x - 3y^2 + 2y - 7 = 0. Now A = 1 and C = -3, so AC < 0: a hyperbola. The y^2 term had been sitting on the other side with a plus sign; moving it changed its sign.
(b) The minus sign in front of (x - 1)^2 looks like a hyperbola’s. Move it across: (x - 1)^2 + (y + 2)^2 = 9. Both squares now carry +1, so A = C, and this is the circle with center (1, -2) and radius 3. Its general form is x^2 + y^2 - 2x + 4y - 4 = 0.
(c) Expand the right side: y^2 + 4x = y^2 - 6y + 9. Subtracting y^2 from both sides removes it completely, leaving 4x + 6y - 9 = 0. Both A and C are zero, so this is not a conic at all but a straight line. An equation can look second degree and not be.
Day 1: a coefficient left open
For each real number k, consider kx^2 + 4y^2 - 6x + 8y - 1 = 0. Which type does the table predict for each value of k?
SolutionHere C = 4 is fixed and A = k varies, so the type depends only on k.
- k = 4 makes A = C: a circle.
- k > 0 with k \ne 4 makes AC = 4k > 0 with A \ne C: an ellipse. For 0 < k < 4 the x^2 coefficient is the smaller one, so the ellipse is wider than it is tall; for k > 4, taller than wide.
- k = 0 removes x^2 while y^2 stays: a parabola, opening sideways.
- k < 0 makes AC = 4k < 0: a hyperbola.
This is what the table promises, “if the equation draws one”. Example 7 completes the square with k still a letter and finds the one value of k where the prediction fails.
Day 2: an ellipse away from the origin
Write 4x^2 + 9y^2 - 16x + 54y + 61 = 0 in standard form. Give the center, the semi-axes and the vertices.
SolutionHere A = 4 and C = 9, so by Day 1 this is an ellipse if it is anything. Group the x terms and the y terms, move the constant across, and factor each leading coefficient out of its group:
Half of -4 is -2, and (-2)^2 = 4. Adding 4 inside the first parentheses adds 4 \cdot 4 = 16 to the left side, because of the factor 4 in front. Half of 6 is 3, and adding 9 inside the second parentheses adds 9 \cdot 9 = 81. Add the same to the right side so the equation stays true:
The right side is positive, so this is a genuine ellipse. Dividing both sides by 36:
Read it as in the ellipse: center (2, -3), horizontal semi-axis a = 3, vertical semi-axis b = 2. The major axis is horizontal, with vertices (-1, -3) and (5, -3); the co-vertices are (2, -1) and (2, -5). If the foci are wanted, c^2 = 9 - 4 = 5 puts them at \displaystyle (2 \pm \sqrt{5}, -3).
Check: the vertex (5, -3) should satisfy the original equation. 4(25) + 9(9) - 16(5) + 54(-3) + 61 equals 100 + 81 - 80 - 162 + 61 = 0.
Day 2: a hyperbola away from the origin
Write 9x^2 - 4y^2 + 36x + 8y - 4 = 0 in standard form. Give the center, the vertices and the asymptotes.
SolutionA = 9 and C = -4 have opposite signs, so this is a hyperbola if it is anything. Group, move the constant, and factor:
Adding 4 inside the first parentheses adds 9 \cdot 4 = 36 to the left. Adding 1 inside the second adds -4 \cdot 1 = -4, since the factor in front is -4. Balance the right side with the same two amounts:
The right side is not zero, so the hyperbola is genuine. Dividing by 36:
The positive term holds x, so the branches open left and right. As in the hyperbola: center (-2, 1), a = 2, b = 3, vertices (-4, 1) and (0, 1). The asymptotes pass through the center with slopes \displaystyle \pm\frac{b}{a} = \pm\frac{3}{2}:
The right vertex (0, 1) is the graph’s y-intercept. Check it in the original equation: 0 - 4 + 0 + 8 - 4 = 0.
Day 2: a parabola with a fraction
Find the vertex of 2x^2 - 12x + 3y + 24 = 0, the direction it opens, and its focus.
SolutionOnly x is squared, so this is a parabola. Keep the x terms on the left and send the rest to the right:
Adding 9 inside the parentheses adds 2 \cdot 9 = 18 to the left, so add 18 to the right:
Factoring -3 out of the right side is the step that exposes k: the vertex is where both sides are zero, at (3, -2). Divide by 2 to reach the standard form of the parabola:
So \displaystyle 4p = -\frac{3}{2} and \displaystyle p = -\frac{3}{8}. The parabola opens downward, because p < 0; its axis is the line x = 3; its focus is \displaystyle \frac{3}{8} below the vertex, at \displaystyle \left(3, -\frac{19}{8}\right), and its directrix is \displaystyle y = -\frac{13}{8}.
Day 2: when the prediction fails
Return to kx^2 + 4y^2 - 6x + 8y - 1 = 0 from Example 3. Complete the square with k as a letter and decide, for every k, whether the equation draws a genuine conic.
SolutionFor k \ne 0, factor k out of the x terms. Half of \displaystyle -\frac{6}{k} is \displaystyle -\frac{3}{k}, so
Substituting both and moving the constants right:
The center is \displaystyle \left(\frac{3}{k}, -1\right), and everything rests on the right side \displaystyle G = 5 + \frac{9}{k}.
- For k > 0, both 5 and \displaystyle \frac{9}{k} are positive, so G > 0 and every such equation is a genuine ellipse. At k = 4 it is the circle \displaystyle \left(x - \frac{3}{4}\right)^2 + (y + 1)^2 = \frac{29}{16}.
- For k < 0, G = 0 exactly when \displaystyle \frac{9}{k} = -5, that is \displaystyle k = -\frac{9}{5}. Then the equation reads \displaystyle 4(y + 1)^2 = \frac{9}{5}\left(x + \frac{5}{3}\right)^2, whose square roots give two lines, \displaystyle y + 1 = \pm\frac{3\sqrt{5}}{10}\left(x + \frac{5}{3}\right), crossing at the center \displaystyle \left(-\frac{5}{3}, -1\right). For every other negative k the hyperbola is genuine. When \displaystyle k < -\frac{9}{5}, G > 0 sits with the positive y term, so the branches open up and down; when \displaystyle -\frac{9}{5} < k < 0, G < 0 sits with the negative x term, and dividing by G makes the x term positive, so they open left and right.
- For k = 0, complete the square in y only: \displaystyle 4(y + 1)^2 = 6x + 5 = 6\left(x + \frac{5}{6}\right), a genuine parabola with vertex \displaystyle \left(-\frac{5}{6}, -1\right), opening right.
So Example 3’s prediction is right for every k except \displaystyle k = -\frac{9}{5}, where the hyperbola collapses to its own pair of asymptotes.
Boundaries
No xy term. Everything here assumes the equation has no xy term. A conic whose axes are neither horizontal nor vertical has one, and then the sign of B^2 - 4AC, where B is the xy coefficient, takes over the job that the sign of AC does here. With B = 0 that quantity is -4AC, so the two tests agree.
A and C not both zero. If both are zero, as in Example 2(c), the equation is linear and draws a line, or nothing.
Degenerate cases. After completing the square, A(x - h)^2 + C(y - k)^2 = G can fail to be a true conic in exactly these ways:
| Squared terms | Right side | Graph |
|---|---|---|
| AC > 0 | G has the sign of A | ellipse or circle |
| AC > 0 | G = 0 | the single point (h, k) |
| AC > 0 | G has the opposite sign | no points at all |
| AC < 0 | G \ne 0 | hyperbola |
| AC < 0 | G = 0 | two lines crossing at (h, k) |
For example, x^2 + 2y^2 - 2x + 8y + 9 = 0 becomes (x - 1)^2 + 2(y + 2)^2 = 0, true only at the point (1, -2): a sum of two squares is zero only when both are. Change the 9 to 10 and the right side becomes -1, which a sum of squares never equals, so there is no graph.
When only x is squared, the equation is A(x - h)^2 + Ey + F' = 0 for some constant F'. With E \ne 0 it is a genuine parabola. With E = 0 it says (x - h)^2 equals a constant: two parallel vertical lines if the constant is positive, the single line x = h if it is zero, and nothing if it is negative. The same holds with the roles of x and y swapped.
The point, the single line and the crossing pair are the cuts of Conic sections from a double cone made by a plane through the cone’s vertex. Two parallel lines are not a slice of a double cone at all; the equation allows them only because a parabola’s one squared variable can lose its partner.
Where this goes next
Reading A and C is the first question to ask of any second degree equation, and completing the square is the move that answers the second, where the curve sits. With the center or vertex in hand, everything in the separate lessons on the ellipse, the hyperbola and the parabola applies: foci, axes, asymptotes, directrix. A single number can also tell the conics apart: the eccentricity of the focus and directrix description, which is between 0 and 1 for an ellipse, 1 for a parabola and above 1 for a hyperbola; a circle is the limit as it shrinks to 0. Later, when the axes are turned and an xy term appears, rotating the coordinate axes removes it and returns the equation to the shape studied here.
Exercises
Day 1: naming the conic
Name the conic each equation draws, and say which coefficients tell you.
- 3x^2 + 3y^2 + 6x - 12y - 1 = 0
- x^2 + 6y^2 - 4x + 12y - 8 = 0
- 5y^2 - 2x^2 + 20y + 4x - 12 = 0
- x^2 - 10x + 4y + 13 = 0
- 7x^2 + 2y^2 - 28x = 0
- y^2 + 3x - 8y + 4 = 0
Day 1: tidy first
Bring each equation to general form, then name its graph.
- (x + 2)^2 + (y - 1)^2 = 2x^2 + 7
- 4x^2 + y(y + 6) = 4(x - 1)^2 + 5
- (x - y)(x + y) = 2x^2 - 9
- In mx^2 + 9y^2 + 2x - 5 = 0, find the value of m that makes the graph a circle and the value that makes it a parabola, and give that circle’s center and radius and that parabola’s vertex.
Day 2: standard form, center and vertex
Write each equation in standard form. Give the center (or the vertex, for a parabola), and the vertices or the asymptotes where the curve has them.
- x^2 + 4y^2 + 6x - 8y - 3 = 0
- 9x^2 + y^2 - 18x + 10y - 11 = 0
- 4x^2 - y^2 - 24x - 4y + 28 = 0
- 16y^2 - 9x^2 - 64y - 54x - 161 = 0
- y^2 - 4y - 6x + 16 = 0
- 3x^2 - 12x + 2y + 10 = 0
- 2x^2 + 2y^2 - 6x + 10y + 1 = 0
- An ellipse has center (-1, 4), horizontal semi-axis 5 and vertical semi-axis 2. Write its equation in general form Ax^2 + Cy^2 + Dx + Ey + F = 0 with integer coefficients.
Day 2: genuine or degenerate
Describe the graph of each equation in the plane.
- x^2 + 3y^2 - 4x - 6y + 7 = 0
- 4x^2 + y^2 + 8x - 2y + 9 = 0
- x^2 - 4y^2 - 2x - 16y - 15 = 0
- x^2 - 2x - 15 = 0
Reasoning
- Kolbrun says completing the square can never change which conic the Day 1 table predicts. Explain why rewriting Ax^2 + Dx as \displaystyle A\left(x + \frac{D}{2A}\right)^2 - \frac{D^2}{4A} leaves the coefficient of x^2 equal to A.
- Orri multiplies a general equation by -1 to make its constant positive, and worries that the type has changed. Explain why it has not, using both the table and the graph.
- Ingolfur claims that if A > 0, C > 0 and F < 0 in Ax^2 + Cy^2 + Dx + Ey + F = 0, the graph is always a genuine ellipse or circle, whatever D and E are. Prove it.
Answers
- A circle, since A = C = 3. Completing the square gives 3(x + 1)^2 + 3(y - 2)^2 = 16.
- An ellipse, since A = 1 and C = 6 have the same sign and differ. It is (x - 2)^2 + 6(y + 1)^2 = 18.
- A hyperbola, since A = -2 and C = 5 have opposite signs. It is 5(y + 2)^2 - 2(x - 1)^2 = 30.
- A parabola, since C = 0: only x is squared. It is (x - 5)^2 = -4(y - 3).
- An ellipse, since A = 7 and C = 2 are both positive and unequal. It is 7(x - 2)^2 + 2y^2 = 28.
- A parabola, since A = 0: only y is squared. It is (y - 4)^2 = -3(x - 4).
- A hyperbola. Expanding gives x^2 + 4x + 4 + y^2 - 2y + 1 = 2x^2 + 7, so -x^2 + y^2 + 4x - 2y - 2 = 0 with A = -1, C = 1. It is (x - 2)^2 - (y - 1)^2 = 1.
- A parabola. Expanding gives 4x^2 + y^2 + 6y = 4x^2 - 8x + 9, so y^2 + 8x + 6y - 9 = 0: the x^2 terms cancel. It is \displaystyle (y + 3)^2 = -8\left(x - \frac{9}{4}\right).
- A circle. The left side is x^2 - y^2, so x^2 - y^2 = 2x^2 - 9 becomes x^2 + y^2 - 9 = 0, the circle x^2 + y^2 = 9 of radius 3 about the origin, despite the minus sign at the start.
- m = 9 gives a circle and m = 0 a parabola. With m = 9: \displaystyle 9\left(x + \frac{1}{9}\right)^2 + 9y^2 = \frac{46}{9}, center \displaystyle \left(-\frac{1}{9}, 0\right), radius \displaystyle \frac{\sqrt{46}}{9}. With m = 0: \displaystyle 9y^2 = -2\left(x - \frac{5}{2}\right), vertex \displaystyle \left(\frac{5}{2}, 0\right).
- \displaystyle \frac{(x + 3)^2}{16} + \frac{(y - 1)^2}{4} = 1, an ellipse with center (-3, 1) and vertices (1, 1) and (-7, 1). The constants added are 9 and 4 \cdot 1 = 4, and 3 + 9 + 4 = 16.
- \displaystyle \frac{(x - 1)^2}{5} + \frac{(y + 5)^2}{45} = 1, an ellipse with center (1, -5) and vertices \displaystyle (1, -5 + 3\sqrt{5}) and \displaystyle (1, -5 - 3\sqrt{5}). Here 9(x - 1)^2 + (y + 5)^2 = 11 + 9 + 25 = 45, and \displaystyle \sqrt{45} = 3\sqrt{5}.
- \displaystyle \frac{(x - 3)^2}{1} - \frac{(y + 2)^2}{4} = 1, a hyperbola with center (3, -2), vertices (2, -2) and (4, -2), asymptotes y + 2 = \pm 2(x - 3). Here 4(x - 3)^2 - (y + 2)^2 = -28 + 36 - 4 = 4.
- \displaystyle \frac{(y - 2)^2}{9} - \frac{(x + 3)^2}{16} = 1, a hyperbola opening up and down with center (-3, 2), vertices (-3, 5) and (-3, -1), asymptotes \displaystyle y - 2 = \pm\frac{3}{4}(x + 3). Here 16(y - 2)^2 - 9(x + 3)^2 = 161 + 64 - 81 = 144.
- (y - 2)^2 = 6(x - 2), a parabola with vertex (2, 2) opening right. Adding 4 to both sides of y^2 - 4y = 6x - 16 gives 6x - 12 on the right.
- \displaystyle (x - 2)^2 = -\frac{2}{3}(y - 1), a parabola with vertex (2, 1) opening downward. Here 3(x - 2)^2 = -2y - 10 + 12 = -2(y - 1).
- \displaystyle \left(x - \frac{3}{2}\right)^2 + \left(y + \frac{5}{2}\right)^2 = 8, a circle with center \displaystyle \left(\frac{3}{2}, -\frac{5}{2}\right) and radius \displaystyle 2\sqrt{2}. Dividing by 2 first, the right side is \displaystyle -\frac{1}{2} + \frac{9}{4} + \frac{25}{4} = 8.
- 4x^2 + 25y^2 + 8x - 200y + 304 = 0. Start from \displaystyle \frac{(x + 1)^2}{25} + \frac{(y - 4)^2}{4} = 1, multiply by 100 to get 4(x + 1)^2 + 25(y - 4)^2 = 100, and expand: 4x^2 + 8x + 4 + 25y^2 - 200y + 400 - 100 = 0.
- The single point (2, 1). Completing the square gives (x - 2)^2 + 3(y - 1)^2 = -7 + 4 + 3 = 0, and a sum of two squares is zero only when both are.
- No points at all. Completing the square gives 4(x + 1)^2 + (y - 1)^2 = -9 + 4 + 1 = -4, and a sum of squares is never negative.
- Two lines crossing at (1, -2): \displaystyle y = \frac{x - 5}{2} and \displaystyle y = \frac{-x - 3}{2}. Completing the square gives (x - 1)^2 - 4(y + 2)^2 = 15 + 1 - 16 = 0, so x - 1 = \pm 2(y + 2).
- Two parallel vertical lines, x = 5 and x = -3. The equation factors as (x - 5)(x + 3) = 0 and says nothing about y, so every point with x = 5 or x = -3 is on the graph.
- Expanding \displaystyle A\left(x + \frac{D}{2A}\right)^2 gives \displaystyle Ax^2 + Dx + \frac{D^2}{4A}: the square contributes exactly one x^2, multiplied by A, and the added number \displaystyle \frac{D}{2A} only reaches the x term and the constant. Subtracting the constant \displaystyle \frac{D^2}{4A} changes no power of x. So the coefficient of x^2 is A before and after, the same is true for y^2 and C, and the type the table reads stays the same.
- Multiplying by -1 replaces A and C by -A and -C. Their product is (-A)(-C) = AC, so its sign is unchanged; A = C holds exactly when -A = -C; and a zero coefficient stays zero. So the table gives the same row. The graph is unchanged too: a point satisfies the new equation exactly when it satisfies the old one, since one side is 0 and 0 \cdot (-1) = 0.
- Completing the square gives A(x - h)^2 + C(y - k)^2 = G with \displaystyle G = \frac{D^2}{4A} + \frac{E^2}{4C} - F. Since A, C > 0, the first two terms of G are at least 0, and -F > 0, so G > 0. Dividing by G gives \displaystyle \frac{(x - h)^2}{G/A} + \frac{(y - k)^2}{G/C} = 1 with both denominators positive: a genuine ellipse, or a circle when A = C. No choice of D and E can make G zero or negative.