Question

The point of intersection of the tangents at the point $P$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and its corresponding point $Q$ on the auxiliary circle meet on the line :

• A. $x=\frac{a}{e}$
• B. $x=0$
• C. $y=0$
• D. None

Answer 1

The correct option is C $y=0$

Let the point $P$ be $(a\cos \theta ,b\sin \theta )$

Equation of tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $P$ is $T=0$

$bx\cos \theta +ay\sin \theta =ab$ ......(i)

Equation of auxiliary circle of ellipse is $x^2+y^2=a^2$

Point corresponding to $P$ on auxiliary circle is $Q(a\cos \theta ,a\sin \theta )$

Equation of tangent to the circle at $Q$ is $T=0$

$x\cos \theta +y\sin \theta =a\Rightarrow x\cos \theta =a-y\sin \theta \Rightarrow x=\frac{a-y\sin \theta }{\cos \theta }$

Substituting $x$ in (i)

$\Rightarrow b\cos \theta \left(\frac{a-y\sin \theta }{\cos \theta }\right)+ay\sin \theta =ab$

$\Rightarrow ab-by\sin \theta +ay\sin \theta =ab$

Thus $(a-b)y\sin \theta =0$

$\Rightarrow y=0$

So, option C is correct.