Browse Questions

# The line $x \cos \alpha + y \sin \alpha =p$ is a tangent to the ellipse. $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1 if \begin{array}{1 1}(A)\;a^2 \cos ^2 \alpha -b^2 \sin ^2 \alpha =p^2 \\(B)\;a^2 \sin ^2 \alpha -b^2 \cos^2 \alpha =p^2 \\(C)\;a^2 \cos^2 \alpha + b^2 \sin ^2 \alpha =p^2 \\(D)\;a^2 \cos^2 \alpha+b^2 \sin ^2 \alpha=p \end{array} Can you answer this question? ## 1 Answer 0 votes We know that line y=mx+c is a tangent to the ellipse. \large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$
$c^2= a^2m^2 +b^2$
In this case $c= \large\frac{-p}{\sin \alpha}$
$m= \large\frac{- \cos \alpha}{\sin \alpha}$
So that given line will be a tangent if
$\large\frac{P^2}{\sin ^2 \alpha} $$=a^2 \large\frac{\cos ^2 \alpha}{\sin ^2 \alpha}$$+b^2$
$p^2=a^2 \cos ^2 \alpha +b^2 \sin ^2 \alpha$
Hence C is the correct answer.