JEE Advanced 2016 Q49 locus

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 $E$. What is the locus of $E$?

Let $P$ be the point $(a,b)$ with $a^2+b^2=1$. The equation of the normal to the circle in $P$ is $y=\frac{b}{a}(x-a)+b$. The equation of the line through $E$ parallel to $RS$ is $y=\frac{b}{a-1}(x-1)+\frac{a}{b}(a-1)+b$. Using some brute force algebra and subtituting $1-a^2$ for $b^2$, I found that $x=\frac{2a}{1+a}$ for the intersection of the two lines. Did I do this correctly? Now, I don’t know how to proceed. I have the $x$-value in $a$ only, but if I substitute this into the equation of one of the lines, I get an answer with $a$ and $b$. Can someone help?

Answer 1

Indeed, you get $x, y$ in terms of $a, b$. If you solve the equations for $a$ and $b$ and substitute that into $a^2+b^2=1$, you get an equation for that locus you're looking for.

Answer 2

Let $P(\cos\theta,\sin\theta)$ so that the equation of the tangent at $P$ is $$y-\sin\theta=-\frac{\cos\theta}{\sin\theta}(x-\cos\theta)$$

$Q$ lies on this line and has $x$ coordinate $1$, so we get, after simplification, the $y$ coordinate of $Q$ which is $$y=\tan\frac{\theta}{2}$$

$E$ lies on the line $OP$ so its coordinates satisfy $$\frac yx=\tan\theta$$

$E$ has the same $y$ coordinate as $Q$. Using the half-angle identity $\tan\theta=\frac{2t}{1-t^2}$, with $t=y$, we can eliminate $\theta$ and obtain the locus of $E$ which is $$y^2=1-2x$$