Find the points on the ellipse `(x^2)/4+(y^2)/9=1` on which the normals are parallel to the line `2x-y=1.`

Options

A. `((9)/(sqrt(10)),(2)/(sqrt(10)))`
B. `(-(9)/(sqrt(10)),(2)/(sqrt(10)))`
C. `(-(2)/(sqrt(10)),(9)/(sqrt(10)))`
D. none of these

Answer

C

Explanation

LEt p( 2 ` cos theta , 3 sin theta)` be the point on the ellipse at which the normal of the normal ar P is
the equation of the normal at P is
`2x sec theta - 3y cosec theta =-5`
if it is parallel to 2x-y=1, then
`(2)/(3) tan theta =2`
`implies tan theta =3`
`implies sin theta =- (3)/( sqrt(10)) and cos theta =(1)/(sqrt(10))`
`or , sin theta =-(3)/(sqrt(10)) and cos theta =-(1)/(sqrt(10))`
hence , the required points are `(+-(2)/(sqrt(10)),+-(9)/(sqrt(10)))`