The line lx+my=n is a normal to the ellipse `x^(2)/a^(2)+y^(2)/b^(2)=1`

Options

A. `(n^(2))/(i n ^(2))+(b^(2))/(l^(2))=((a^(2)-b^(2))^(2))/(n^(2))`
B. `(a^(2))/(l ^(2))+(b^(2))/(m^(2))=((a^(2)-b^(2))^(2))/(n^(2))`
C. `(n^(2))/(l ^(2))+(b^(2))/(m^(2))=((a^(2)-b^(2))^(2))/(n^(2))`
D. none of these

Answer

B

Explanation

Let the line lx + my + n =0 be normal to the ellipse at the point `( a cos theta , b sin theta ).` then the equation of the normal at `( a cos theta , b sin theta )` is
`ax sec theta - by cosec theta = a^(2) -b^(2) `
if this is to be same as the line `lx+ my+ n- 0 ` then
`( asec theta )/(l) =-(b" cosec" theta)/( m) =(a^(2)-b^(2))/(-n)`
`implies cos theta =-( an)/( l(a^(2)-b^(2))) and sin theta =(an)/(" in "(a^(2)-b^(2)))`
`implies (a^(2)n^(2))/(l^(2)(a^(2)-b^(2))^(2))+(b^(2)n^(2))/("in"^(2)(a^(2)-b^(2))^(2))=1`
[ Squaring and adding ]
`implies (a^(2))/(l^(2))+(b^(2))/(m^(2))=((a^(2)-b^(2))^(2))/(n^(2))`