Question

Find an equation for the ellipse with vertices at (-3, 4) and (15, 4); focus at (13, 4)

Answer 1

Vertices of the ellipse are (-3, 4) and (15, 4) and focus at (13, 4)

The y coordinate of vertices and focus are same so the ellipse is horizontal.

Standard form of horizontal ellipse is [(x - h)²]/a² + [(y - k)²]/b² = 1

Where a² > b²

(h, k) is center.

Vertices are (h + a, k),(h - a, k)

Focus (h + c , k ), (h - c , k )

 

(h + a, k) = (- 3, 4)

(h - a, k) = (15, 4)

h + a = - 3 ---> (1)

h - a = 15 ---> (2)

Add the equations (1) and (2).

2h = 12

h = 6

Substitute the h value in equation (1).

6 + a = - 3

a = - 3 - 6

a = - 9

Center of ellipse (h, k) = (6, 4)

 

Focus (h + c , k ) = (13, 4)

h + c = 13

Substitute the h value in above equation.

6 + c = 13

c = 13 - 6

c = 7

c is distance from center to each focus.

c = √(a² - b² )

c² = a² - b²

b² = a² - c²

b = √(a² - c²)

b = √[(- 9)² - 7²]

b = √32

Substitute the a,b and center values in standard form [(x - h)²]/a² + [(y - k)²]/b² = 1

Required equation [(x - 6)²]/81 + [(y - 4)²]/32 = 1.