Question

Let $RS$ be the diameter of the circle $x^2+y^2=1,$ where $S$ is the point $(1,0).$ Let $P$ be a variable point $($other than $R$ and $S)$ on the circle and tangents to the circle at $S$ and $P$ meet at the point $Q.$ Then normal to the circle at $P$ intersects a line drawn through $Q$ parallel to $RS$ at point $E.$ Then the locus of $E$ passes through the point(s)?

• A. $(\frac{1}{3},-\frac{1}{\sqrt{3}})$
• B. $(\frac{1}{4},\frac{1}{2})$
• C. $(\frac{1}{3},\frac{1}{\sqrt{3}})$
• D. $(\frac{1}{4},-\frac{1}{2})$

Answer 1

Answer: (A), (C)

$(\frac{1}{3},\frac{1}{\sqrt{3}})$



Tangent at $P:x\cos \theta +y\sin \theta =1$
At $x=1$ we get $Q:(1,\frac{1-\cos \theta }{\sin \theta })$

Normal at $P:y=\left(\tan \theta \right)x$
Equation of line from point $Q$ and parallel to $RS$ is
$L:y=\frac{1-\cos \theta }{\sin \theta }$
$\Rightarrow y=\frac{2{\sin }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}=\tan \frac{\theta }{2}$
Let point of intersection of line $L$ and normal at $P$ be $(h,k).$ Then
$\tan \theta =\frac{k}{h}$ and $\tan \frac{\theta }{2}=k$
From$\frac{k}{h}=\frac{2\tan \frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}$
$\Rightarrow \frac{k}{h}=\frac{2k}{1-k^2}$
$\Rightarrow k=0$ or $1-k^2=2h$
But $y=0$ is not possible then required locus is $y^2=1-2x$
$\therefore$ points $(\frac{1}{3},\pm \frac{1}{\sqrt{3}})$ lie on the locus.