Question

If the tangent at any point $P$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the tangents at the vertices $A$ and $A^{\prime }$ in $L$ and $L^{\prime }$ respectively, then $AL\cdot A^{\prime }{L}^{\prime }$ is equal to

• A. $a+b$
• B. $a^2+b^2$
• C. $a^2$
• D. $b^2$

Answer 1

Answer: (D)

Let $P(a\cos \theta ,b\sin \theta )$ be any point on the ellipse. Then, equation of the tangent at $P$ is $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$
It cuts the lines $x=a$ and $x=-a$ $L\left(a,\frac{b(1-\cos \theta )}{\sin \theta }\right)$ and $L^{\prime }\left(-a,\frac{b(1+\cos \theta )}{\sin \theta }\right)$ respectively.
Since, $A$ and $A^{\prime }$ are the vertices of given ellipse.
Therefore, coordinates $A(a,0)$ and $B(0,-a)$ .
$\therefore$ $AL=\frac{b(1-\cos \theta )}{\sin \theta }$ and $A{L}^{\prime }=\frac{b(1+\cos \theta )}{\sin \theta }$