Ellipses and Hyperbolas

Standard forms and key features of ellipses and hyperbolas

Ellipses and Hyperbolas

Ellipses

An ellipse is the set of all points where the sum of distances to two foci is constant.

Standard Form (Horizontal Major Axis)

(xh)2a2+(yk)2b2=1,a>b\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1, \quad a > b

Standard Form (Vertical Major Axis)

(xh)2b2+(yk)2a2=1,a>b\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1, \quad a > b

Where:

  • (h,k)(h, k) = center
  • aa = semi-major axis (larger value)
  • bb = semi-minor axis (smaller value)
  • cc = distance from center to focus

Key relationship: c2=a2b2c^2 = a^2 - b^2

Key Features of Ellipse

For horizontal major axis (xh)2a2+(yk)2b2=1\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1:

  • Center: (h,k)(h, k)
  • Vertices (endpoints of major axis): (h±a,k)(h \pm a, k)
  • Co-vertices (endpoints of minor axis): (h,k±b)(h, k \pm b)
  • Foci: (h±c,k)(h \pm c, k) where c2=a2b2c^2 = a^2 - b^2
  • Major axis length: 2a2a
  • Minor axis length: 2b2b

For vertical major axis: Swap the roles (vertices on vertical axis)

Eccentricity

e=ca,0<e<1e = \frac{c}{a}, \quad 0 < e < 1

  • ee close to 00: nearly circular
  • ee close to 11: very elongated

Hyperbolas

A hyperbola is the set of all points where the difference of distances to two foci is constant.

Standard Form (Horizontal Transverse Axis)

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

Opens left and right.

Standard Form (Vertical Transverse Axis)

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

Opens up and down.

Where:

  • (h,k)(h, k) = center
  • aa = distance from center to vertex
  • bb = determines spread of branches
  • cc = distance from center to focus

Key relationship: c2=a2+b2c^2 = a^2 + b^2 (note: plus, not minus!)

Key Features of Hyperbola

For horizontal transverse axis (xh)2a2(yk)2b2=1\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1:

  • Center: (h,k)(h, k)
  • Vertices: (h±a,k)(h \pm a, k)
  • Foci: (h±c,k)(h \pm c, k) where c2=a2+b2c^2 = a^2 + b^2
  • Asymptotes: yk=±ba(xh)y - k = \pm\frac{b}{a}(x - h)

For vertical transverse axis: Vertices and foci on vertical axis, asymptotes: yk=±ab(xh)y - k = \pm\frac{a}{b}(x - h)

Asymptotes

The branches of a hyperbola approach (but never touch) the asymptotes.

Rectangle method:

  1. Draw rectangle with vertices at (h±a,k±b)(h \pm a, k \pm b)
  2. Draw diagonals of rectangle
  3. These diagonals are the asymptotes

Identifying Conic Sections

From the equation Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 (with B=0B = 0):

  • Circle: A=CA = C (and same sign)
  • Ellipse: ACA \neq C but same sign
  • Parabola: Either A=0A = 0 or C=0C = 0 (but not both)
  • Hyperbola: AA and CC have opposite signs

Summary Table

| Conic | Standard Form | cc relationship | Key feature | |-------|---------------|------------------|-------------| | Circle | (xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2 | N/A | All points distance rr from center | | Ellipse | (xh)2a2+(yk)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 | c2=a2b2c^2 = a^2 - b^2 | Sum of distances to foci = 2a2a | | Hyperbola | (xh)2a2(yk)2b2=1\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 | c2=a2+b2c^2 = a^2 + b^2 | Difference of distances to foci = 2a2a | | Parabola | (xh)2=4p(yk)(x-h)^2 = 4p(y-k) | N/A | Distance to focus = distance to directrix |

📚 Practice Problems

1Problem 1easy

Question:

Find the vertices, co-vertices, and foci of the ellipse (x2)225+(y+1)29=1\frac{(x - 2)^2}{25} + \frac{(y + 1)^2}{9} = 1.

💡 Show Solution

Identify the ellipse characteristics:

Step 1: Determine orientation:

  • Form: (xh)2a2+(yk)2b2=1\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1
  • a2=25>b2=9a^2 = 25 > b^2 = 9 → horizontal major axis
  • a=5a = 5, b=3b = 3
  • Center: (h,k)=(2,1)(h, k) = (2, -1)

Step 2: Find vertices (endpoints of major axis): Vertices are aa units left and right of center: (2±5,1)=(7,1) and (3,1)(2 \pm 5, -1) = (7, -1) \text{ and } (-3, -1)

Step 3: Find co-vertices (endpoints of minor axis): Co-vertices are bb units up and down from center: (2,1±3)=(2,2) and (2,4)(2, -1 \pm 3) = (2, 2) \text{ and } (2, -4)

Step 4: Find foci using c2=a2b2c^2 = a^2 - b^2: c2=259=16c^2 = 25 - 9 = 16 c=4c = 4

Foci are cc units left and right of center: (2±4,1)=(6,1) and (2,1)(2 \pm 4, -1) = (6, -1) \text{ and } (-2, -1)

Answers:

  • Center: (2,1)(2, -1)
  • Vertices: (7,1)(7, -1) and (3,1)(-3, -1)
  • Co-vertices: (2,2)(2, 2) and (2,4)(2, -4)
  • Foci: (6,1)(6, -1) and (2,1)(-2, -1)
  • Major axis length: 2a=102a = 10
  • Minor axis length: 2b=62b = 6

2Problem 2medium

Question:

Find the center, vertices, foci, and asymptotes of the hyperbola (y+3)216(x1)29=1\frac{(y + 3)^2}{16} - \frac{(x - 1)^2}{9} = 1.

💡 Show Solution

Identify the hyperbola characteristics:

Step 1: Determine orientation:

  • yy term is positive → vertical transverse axis (opens up and down)
  • a2=16a^2 = 16, so a=4a = 4
  • b2=9b^2 = 9, so b=3b = 3
  • Center: (h,k)=(1,3)(h, k) = (1, -3)

Step 2: Find vertices: Vertices are aa units above and below center: (1,3±4)=(1,1) and (1,7)(1, -3 \pm 4) = (1, 1) \text{ and } (1, -7)

Step 3: Find foci using c2=a2+b2c^2 = a^2 + b^2: c2=16+9=25c^2 = 16 + 9 = 25 c=5c = 5

Foci are cc units above and below center: (1,3±5)=(1,2) and (1,8)(1, -3 \pm 5) = (1, 2) \text{ and } (1, -8)

Step 4: Find asymptotes: For vertical hyperbola: yk=±ab(xh)y - k = \pm\frac{a}{b}(x - h) y(3)=±43(x1)y - (-3) = \pm\frac{4}{3}(x - 1) y+3=±43(x1)y + 3 = \pm\frac{4}{3}(x - 1)

Two asymptotes: y=43(x1)3=43x133y = \frac{4}{3}(x - 1) - 3 = \frac{4}{3}x - \frac{13}{3} y=43(x1)3=43x53y = -\frac{4}{3}(x - 1) - 3 = -\frac{4}{3}x - \frac{5}{3}

Answers:

  • Center: (1,3)(1, -3)
  • Vertices: (1,1)(1, 1) and (1,7)(1, -7)
  • Foci: (1,2)(1, 2) and (1,8)(1, -8)
  • Asymptotes: y=43x133y = \frac{4}{3}x - \frac{13}{3} and y=43x53y = -\frac{4}{3}x - \frac{5}{3}

3Problem 3hard

Question:

Write the equation of an ellipse with center at the origin, a focus at (0,3)(0, 3), and a vertex at (0,5)(0, 5).

💡 Show Solution

Determine ellipse characteristics:

Step 1: Analyze given information:

  • Center: (0,0)(0, 0)
  • Focus: (0,3)(0, 3) → on yy-axis → vertical major axis
  • Vertex: (0,5)(0, 5) → also on yy-axis ✓

Step 2: Find aa and cc:

  • Distance from center to vertex: a=5a = 5
  • Distance from center to focus: c=3c = 3

Step 3: Use c2=a2b2c^2 = a^2 - b^2 to find bb: 32=52b23^2 = 5^2 - b^2 9=25b29 = 25 - b^2 b2=16b^2 = 16 b=4b = 4

Step 4: Write equation (vertical major axis with center at origin): x2b2+y2a2=1\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 x216+y225=1\frac{x^2}{16} + \frac{y^2}{25} = 1

Step 5: Verify:

  • Center: (0,0)(0, 0)
  • a=5a = 5, so vertices: (0,±5)(0, \pm 5)
  • c=3c = 3, so foci: (0,±3)(0, \pm 3)
  • b=4b = 4, so co-vertices: (±4,0)(\pm 4, 0)

Answer: x216+y225=1\frac{x^2}{16} + \frac{y^2}{25} = 1