if tangents are drawn to the ellipse `x^(2)+2y^(2)=2` all points on theellipse other its four vertices then the mid-points of the tangents intercepted between the coorinate axs lie on the curve

Options

A. `(x^(2))/(4)+(y^(2))/(2)=1`
B. `(x^(2))/(4x^(2))+(y^(2))/(2y^(2))=1`
C. `(x^(2))/(2)+(y^(2))/(4)=1`
D. `(x^(2))/(2x^(2))+(y^(2))/(4y^(2))=1`

Answer

D

Explanation

Given equation of ellipse is `x^(2)+ 2y^(2)=2` which can be written as `(x^(2))/(2)+(y^(2))/(1)=1`
Let P be point on the ellipse , other than its four vertices , then the parametric coordinates of P be `(sqrt(2) cos theta , sin theta)`

Now , the equation of tangent at P is
`( x sqrt(2) cos theta)/(2) +( y sin theta)/(1) =1`
` [:'` equation of tangent at `( x_(1),y_(1)_)` is given by T=0 `implies ( x x _(1))/(a^(2))+(yy_(1))/(b^(2))=1`]
`implies ( x) /( sqrt(2)sec theta)+ ( y) /(cosec theta )=1`
`therefore A ( sqrt(2)sextheta , 0) and B (0, cossec theta )`
`Let " mid-point of " AB be R(h,k), `then
` h= ( sqrt(2)sec theta)/( 2) and k= ( cosec theta )/(2) `
`2h= sqrt(2) sec theta and 2K = cosec theta `
`implies cos theta =(1)/( sqrt(2h)) and sin theta `
we know that , `cos^(2) theta + sin ^(2) sin ^(2) theta =1 `
` therefore (1) /( 2h^(2))+(1)/(4k^(2))1`
= so locus of `( h. k) ` is `(1) /( 2x^(2))+ (1)/(4y^(2))=1`