Question

Given that 9y^2+25y^2=225,find the covertices and vertices,foci and length of the latus rectum?

Answer 1

Coordinates of the vertices `(-5,0);(5,0)`
Coordinates of the covertices `(0,3);(0,-3)`
coordinates of the foci `(-4,0);(4,0)`
Latus Rectum of the ellipse `=18/5`

Explanation:

There is a mistake in the problem

The problem shall be `9x^2+25y^2=225`
[it cannot be `9y^2+25y^2=225`]

It is an ellipse.
The standard form of an ellipse is

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

Let us divide both sides of the given equation to have it in the standard form

`(9x^2)/225+(25y^2)/225=225/225`

`x^2/25+y^2/9=1`

Since `a^2` is greater than `b^2`, the major axis of ellipse is along the X-axis and minor axis along the Y-axis and its centre is at `(0,0)`
Then -
Vertices

`a^2=25`
`a=5`

`(-5,0);(5,0)`

Co-vertices

`b^2=9`
`b=3`

`(0,-3);(0,3)`

coordinates of the foci

`c^2+b^2=a^2`
`c^2+9=25`
`c^2=25-9=16`
`c=4`
Then `(-4,0);(4,0)`

Latus Rectum of the ellipse `=(2b^2)/a=(2xx 9)/5=18/5`