Question

How do you find the standard form of the equation of the ellipse given the properties foci `(0,+-5)`, vertices `(0, +-8)`?

Answer 1

The equation of the ellipse is `y^2/64+x^2/39=1`

Explanation:

The equation of an ellipse with major vertical axis is

`(y-k)^2/a^2+(x-h)^2/b^2=1`

The center( symmetric wrt the foci and the vertices) of the ellipse is

`C=(h,k)=(0,0)`

Therefore,

`a=8`

`c=5`

`b^2=(a^2-c^2)=(64-25)=sqrt39`

The equation of the ellipse is

`y^2/64+x^2/39=1`

graph{(y^2/64+x^2/39-1)=0 [-17.3, 18.75, -8.67, 9.35]}

Answer 2

The standard equation of vertical ellipse is `x^2/39+y^2/64=1`

Explanation:

The vertices and foci are on the y axis at

`V(0, 8), V'(0, -8) and F(0, 5) , F'(0, -5)`

Semi major axis is `a=8` and focus `c=5` from the center

`(0,0)`. This is vertical ellipsce of which the equation is

`x^2/b^2+y^2/a^2=1 or x^2/b^2+y^2/8^2=1`

`c=5` is the distance from the center to a focus. The relation of

`c, a, b` is `c^2 = a^2 - b^2:. 5^2=8^2-b^2` or

`b^2=64-25=39 :. b= sqrt 39 ~~6.245`, therefore, the equation

of vertical ellipse is `x^2/39+y^2/64=1`

graph{x^2/39+y^2/64=1 [-20, 20, -10, 10]} [Ans]