Question

Let the tangent at $P$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ cuts the major axis in $T$ and $PN$ is the perpendicular to $x$ axis. If $C$ is the centre of the ellipse then $CT\cdot CN$ will be equal to:

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

Answer 1

The correct option is D $a^2$

Given the equation of the ellipse is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Center of ellipse is $C\equiv (0,0)$

Now, equation of tangent to ellipse at $P(a\cos \theta ,b\sin \theta )$ is

$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$

Since, the tangent meet $x-$axis at $T$ i.e. $y=0$

$\Rightarrow x=a\sec \theta$

So coordinates of $T$ are $(a\sec \theta ,0)$

$\Rightarrow CT=a\sec \theta$

Since, $PN$ is perpendicular to $x-$axis

So, coordinates of $N$ are $(a\cos \theta ,0)$

$\Rightarrow CN=a\cos \theta$

$CN.CT=\left(a\cos \theta \right)\left(a\sec \theta \right)=a^2$

Hence, option D.