Question

9x2+4y2 = 36 The foci are located at:

Answer 1

The foci are located at
`(0,\pm √(5))`

Explanation:

The standard equation of an ellipse with a vertical major axis is
`(x^2)/(b^2)+(y^2)/(a^2)=1` where
`a^2\:>\:b^2`.

The given equation is
`9x^2+4y^2=36`.

To obtain the standard form, we must divide through by 36.

`(9x^2)/(36)+(4y^2)/(36)=(36)/(36)`

We simplify by canceling out the common factors to obtain;

`(x^2)/(4) +(y^2)/(9)=1`

By comparing this equation to

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

We have
`a^2=9,b^2=4`.

To find the foci, we use the relation:
`a^2-b^2=c^2`

This implies that:

`9-4=c^2`

`c^2=5`

`c=\pm√(5)`

The foci are located at
`(0,\pm c)`

Therefore the foci are
`(0,\pm √(5))`

Or

`(0,-√(5))` and
`(0,√(5))`