if the line `xcosalpha+ysinalpha=p` is normal to the ellipse `x^2/a^2+y^2/b^2=1` then

Options

A. `p^(2)(a^(2)cos^(2)alpha+b^(2)sin^(2)alpha)=a^(2)-b^(2)`
B. `p^(2)(a^(2)cos^(2)alpha+b^(2)sin^(2)alpha)=(a^(2)-b^(2))`
C. `p^(2)(a^(2)sec^(2)alpha+b^(2)cosec^(2)alpha)=a^(2)-b^(2)`
D. `p^(2)(a^(2)sec^(2)alpha+b^(2)cosec^(2)alpha)=(a^(2)-b^(2))^(2)`

Answer

D

Explanation

The equation of any normal to `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` is `ax sec phi - by "cosec" phi= a^(2)-b^(2) " "...(i)`
The straight line `x cos a + y sin alpha` = p will a normal to the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`, if Eq. (i) and `x cos alpha+ y sin alpha=p` represent the same line.
`:. (a sec phi)/(cos alpha)=(-b "cosec" phi)/(sin alpha)=(a^(2)-b^(2))/(p)`
`rArr cos phi =(ap)/((a^(2)-b^(2))cos alpha)`
`sin phi=(-bp)/((a^(2)-b^(2)) sin alpha)`
`:. sin^(2) phi + cos^(2) phi=1`
`rArr (b^(2)p^(2))/((a^(2)-b^(2))^(2) sin^(2) alpha)+(a^(2)p^(2))/((a^(2)-b^(2))^(2)cos^(2) alpha)=1`
`rArr p^(2)(b^(2) "cosec"^(2) alpha+a^(2) sec^(2) alpha)=(alpha^(2)-b^(2))^(2)`