Question

Which of the following is a foci of x^2/9 +y^2/4=1

(5,0)
(0,5)
(13,0)

Answer 1

Coordinates of foci are
`(-√(5),0)\boldsymbol{\texttt{ and }}(√(5),0)`

Not given in option.

Explanation:

For the ellipse

`(x^2)/(a^2)+(y^2)/(b^2)=1` foci are at (-c,0) and (c,0), where
`c=√(a^2-b^2)`

Comparing

`(x^2)/(9)+(y^2)/(4)=1\boldsymbol{\texttt{ with }}(x^2)/(a^2)+(y^2)/(b^2)=1`

a² = 9 and b² = 4

a = 3 and b =2

So
`c=√(3^2-2^2)=√(5)`

Coordinates of foci are
`(-√(5),0)\boldsymbol{\texttt{ and }}(√(5),0)`

Not given in option.

Answer 2

we have that
x²/9 +y²/4=1
a=3
b=2
Find c, the distance from the center to a focus.
c²=a²-b²------> c²=9-4----> c²=5-----> c=√5
(h,k)----> (0,0)

The first focus------> (h+c,k)-----> (√5,0)
The second focus------> (h-c,k)-----> (-√5,0)


the answer is
(√5,0)