09 Conic Sections

A structured guide to identifying, converting, graphing, and comparing parabolas, ellipses, and hyperbolas using standard equations and geometric properties.

Recognizing the Three Conics

A conic section is a curve formed by intersecting a plane with a double cone. In analytic geometry, conics are represented by second-degree equations involving xx and yy, but each principal conic also has a geometric distance definition.

  • A has equal distances from a and a .

  • An has a constant sum of distances from two foci.

  • A has a constant absolute difference of distances from two foci.

For a nonrotated equation, inspect the squared terms first:

  • Only one variable is squared: .

  • Both variables are squared with the same sign but different coefficients: .

  • The squared terms have opposite signs: .

  • The squared terms have equal positive coefficients: circle, which is a special .

These quick classifications apply to equations without an xyxy term. An equation containing xyxy may describe a rotated conic and generally requires a rotation-of-axes method or technology for a complete graph.

Takeaway: Identify the conic from the squared terms, then convert to standard form to reveal its graphing features.

Parabolas: Vertex, , and

A is identified by its vertex, axis of symmetry, opening direction, , and . Its standard forms are

(y−k)2=4p(x−h)(y-k)^2=4p(x-h)

for a horizontal and

(x−h)2=4p(y−k)(x-h)^2=4p(y-k)

for a vertical .

The vertex is (h,k)(h,k). For a horizontal , the is (h+p,k)(h+p,k), the is x=h−px=h-p, and the axis of symmetry is y=ky=k. For a vertical , the is (h,k+p)(h,k+p), the is y=k−py=k-p, and the axis of symmetry is x=hx=h. The sign of pp gives the opening direction.

To graph a :

  1. Identify the vertex (h,k)(h,k).

  2. Determine whether the equation is horizontal or vertical.

  3. Use the sign of pp to determine the opening direction.

  4. Plot the and when useful.

  5. Plot symmetric points on both sides of the axis.

For example, consider

(x−2)2=8(y+1).(x-2)^2=8(y+1).

Because 8=4p8=4p, p=2p=2. Thus the vertex is (2,−1)(2,-1), the opens upward, the is (2,1)(2,1), and the is y=−3y=-3. Setting x−2=±2x-2=\pm2 gives the symmetric points (0,−12)(0,-\frac{1}{2}) and (4,−12)(4,-\frac{1}{2}).

Takeaway: In a , the squared expression identifies the axis direction, while pp determines the , , and opening.

Ellipses: Axes, Vertices, and Foci

An is a closed curve with a center, vertices, co-vertices, major axis, minor axis, and foci. Its standard forms are

(x−h)2a2+(y−k)2b2=1,a>b>0\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1, \qquad a>b>0

for a horizontal major axis, and

(x−h)2b2+(y−k)2a2=1,a>b>0\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1, \qquad a>b>0

for a vertical major axis.

The center is (h,k)(h,k). The larger denominator is a2a^2, so it identifies the major-axis direction. The foci satisfy

c2=a2−b2.c^2=a^2-b^2.

For a horizontal , the vertices are (h±a,k)(h\pm a,k), the co-vertices are (h,k±b)(h,k\pm b), and the foci are (h±c,k)(h\pm c,k). For a vertical , the vertices are (h,k±a)(h,k\pm a), the co-vertices are (h±b,k)(h\pm b,k), and the foci are (h,k±c)(h,k\pm c). The major and minor axis lengths are 2a2a and 2b2b, respectively.

For example, consider

(x−1)225+(y+2)29=1.\frac{(x-1)^2}{25}+\frac{(y+2)^2}{9}=1.

The center is (1,−2)(1,-2), with a=5a=5 and b=3b=3. Since the larger denominator is under the xx-term, the major axis is horizontal. The vertices are (−4,−2)(-4,-2) and (6,−2)(6,-2), while the co-vertices are (1,−5)(1,-5) and (1,1)(1,1). Since c2=25−9=16c^2=25-9=16, c=4c=4, so the foci are (−3,−2)(-3,-2) and (5,−2)(5,-2).

Takeaway: Read the center from the translations, use the larger denominator to find the major-axis direction, and use c2=a2−b2c^2=a^2-b^2 to locate the foci.

Hyperbolas: Branches and Asymptotes

A has two separate branches. Its standard horizontal form is

(x−h)2a2−(y−k)2b2=1,\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1,

and its standard vertical form is

(y−k)2a2−(x−h)2b2=1.\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1.

The positive squared term identifies the direction of opening. For a horizontal , the vertices are (h±a,k)(h\pm a,k), the foci are (h±c,k)(h\pm c,k), and the asymptotes are

y−k=±ba(x−h).y-k=\pm\frac{b}{a}(x-h).

For a vertical , the vertices are (h,k±a)(h,k\pm a), the foci are (h,k±c)(h,k\pm c), and the asymptotes are

y−k=±ab(x−h).y-k=\pm\frac{a}{b}(x-h).

The focal relationship is

c2=a2+b2.c^2=a^2+b^2.

For example, consider

(x+3)216−(y−1)29=1.\frac{(x+3)^2}{16}-\frac{(y-1)^2}{9}=1.

The center is (−3,1)(-3,1), and the positive xx-term shows that the branches open left and right. Here, a=4a=4 and b=3b=3, so the vertices are (−7,1)(-7,1) and (1,1)(1,1). The asymptotes are

y−1=±34(x+3).y-1=\pm\frac{3}{4}(x+3).

Because c2=42+32=25c^2=4^2+3^2=25, c=5c=5, and the foci are (−8,1)(-8,1) and (2,1)(2,1).

To sketch a , plot the center, construct a rectangle with half-widths aa and bb, draw its diagonals as asymptotes, and sketch branches opening toward the vertices.

Takeaway: The positive term determines the opening direction, the vertices are aa units from the center, and the asymptotes guide the two branches.

Converting General Equations to Standard Form

Many conic equations become easier to interpret after rearranging terms and . The goal is to reach a standard equation whose translations and coefficients directly reveal the graph.

For a , isolate the quadratic expression and complete the square in the squared variable. For example,

y2−6y−4x+1=0y^2-6y-4x+1=0

can be rearranged as

y2−6y=4x−1.y^2-6y=4x-1.

Add 99 to both sides:

y2−6y+9=4x+8,y^2-6y+9=4x+8,

which gives

(y−3)2=4(x+2).(y-3)^2=4(x+2).

Therefore, the vertex is (−2,3)(-2,3), the opens right, and p=1p=1.

For an , group the quadratic terms, complete the square in both variables, and divide so that the right-hand side is 11. For example,

9x2+4y2−18x+16y−11=09x^2+4y^2-18x+16y-11=0

becomes

9(x−1)2+4(y+2)2=36,9(x-1)^2+4(y+2)^2=36,

and therefore

(x−1)24+(y+2)29=1.\frac{(x-1)^2}{4}+\frac{(y+2)^2}{9}=1.

This identifies an centered at (1,−2)(1,-2) with a vertical major axis, a=3a=3, and b=2b=2.

For a general nonrotated equation

Ax2+Cy2+Dx+Ey+F=0,Ax^2+Cy^2+Dx+Ey+F=0,

compare the signs and coefficients of the squared terms before doing detailed algebra. If the equation includes an xyxy term,

Ax2+Bxy+Cy2+Dx+Ey+F=0,Ax^2+Bxy+Cy^2+Dx+Ey+F=0,

the conic may be rotated, so ordinary completing-the-square steps may not provide the full orientation.

Takeaway: Classify first, complete the square second, and divide into standard form when possible. Standard form exposes the center or vertex, orientation, scale, foci, , or asymptotes.

Comparing and Graphing Conics

The three principal noncircular conics can be compared by their shape, distance property, and main graphing features.

    • Branches: one.

    • Distance property: points are equally distant from a and a .

    • Center: usually none.

    • Main graphing features: vertex and axis of symmetry.

    • Parameter relationship: uses pp.

    • Branches: one closed curve.

    • Distance property: the sum of distances to two foci is constant.

    • Center: yes.

    • Main graphing features: center, vertices, and co-vertices.

    • Parameter relationship: c2=a2−b2c^2=a^2-b^2.

    • Branches: two.

    • Distance property: the absolute difference of distances to two foci is constant.

    • Center: yes.

    • Main graphing features: center, vertices, and asymptotes.

    • Parameter relationship: c2=a2+b2c^2=a^2+b^2.

A reliable workflow is:

  1. Inspect the squared terms and classify the conic.

  2. Rearrange the equation and complete the square.

  3. Convert to standard form.

  4. Read the vertex or center and determine the orientation.

  5. Calculate foci, , co-vertices, or asymptotes as appropriate.

  6. Plot the key points and sketch the curve symmetrically.

Final takeaway: Standard form is the bridge between an algebraic equation and a geometric graph. The signs, coefficients, translations, and parameter relationships together determine the conic's essential structure.