Find the equation of the ellipse whose axes are along the coordinate axes, vertices are `(0,+-10)` and eccentricity`e=4//5` .

Options

A. `36x^(2)+100y^(2)=3600`
B. `36x^(2)+100y^(2)=1`
C. `100x^(2)+36y^(2)=3600`
D. `100x^(2)+36y^(2)=1`

Answer

C

Explanation

Let the equation of the required be
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
since the vertices of the ellipse are on y-axes .so the coordinates of the vertices are `(0,+- b)`
`therefore b=10`
the eccentricity e is given by
`a^(2)=b^(2)(1-e^(2))implies a^(2)=100(1-(16)/(25))=36`
substituting the values of `a^(2) and b^(2)` in (i) , we obtain
`(x^(2))/(36)+(y^(2))/(100)=1` as the equation of the required ellipse .