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 $Q$. The 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)

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

We can see the line on which point $E$ lies is having equation $y=\tan \theta x$ now as we can see the $y$ co-ordinate of $E$ is $\frac{1-\cos \theta }{\sin \theta }$ subsituting in the equation we get the co-ordinate of $E$ as $\left(\frac{1-\cos \theta }{\sin \theta \tan \theta }\right),\left(\frac{1-\cos \theta }{\sin \theta }\right)$
$E(\left(\frac{1-cos\theta }{sin\theta tan\theta }\right),\left(\frac{1-cos\theta }{sin\theta }\right))\Rightarrow E(\frac{tan\frac{\theta }{2}}{tan\theta },tan\frac{\theta }{2})$
Let $h=\frac{tan\frac{\theta }{2}}{tan\theta }$ and $k=\tan \frac{\theta }{2}$
$\therefore h=\frac{k}{tan\theta }\therefore tan\frac{\theta }{2}=\frac{k}{n}$
$\frac{2tan\frac{\theta }{2}}{1-ta{n}^2\frac{\theta }{2}}=\frac{k}{h}\Rightarrow \left(\frac{2k}{1-k^2}\right)=\frac{k}{h}$

$\therefore 2xy=y(1-y^2)$, replacing $h,k$ by $x,y$