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)

The correct options are
A

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

C

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

Given, RS is the diameter of $x^2+y^2=1$
Here, equation of the tangent at $p(cos\theta ,sin\theta )$ is $xcos\theta +ysin\theta =1$


This tangent intersects with the tangent x=1
$\Rightarrow y=\frac{1-cos\theta }{sin\theta }\therefore Q(1,\frac{1-cos\theta }{sin\theta })$
$\therefore$ Equation of the line through Q parallel to RS is
$y=\frac{1-cos\theta }{sin\theta }=\frac{2si{n}^2\frac{\theta }{2}}{2sin\frac{\theta }{2}cos\frac{\theta }{2}}=tan\frac{\theta }{2}\dots \dots \left(i\right)$
Normal at P: $y=\frac{sin\theta }{cos\theta }.x\Rightarrow y=xtan\theta \dots \dots \left(ii\right)$
Let their point of intersection be (h, k)
Then $k=tan\frac{\theta }{2}andk=htan\theta \therefore k=h\left(\frac{2tan\frac{\theta }{2}}{1-ta{n}^2\frac{\theta }{2}}\right)\Rightarrow k=\frac{2h.k}{1-k^2}\Rightarrow k(1-k^2)=2hk$
$\therefore$ Locus for point E: $2x=1-y^2\dots \dots \left(iii\right)$
When $x=\frac{1}{3}$, then
$1-y^2=\frac{2}{3}\Rightarrow y^2=1-\frac{2}{3}\Rightarrow y=\pm \frac{1}{\sqrt{3}}$
$\therefore (\frac{1}{3},\pm \frac{1}{\sqrt{3}})$ satisfy $2x=1-y^2$
When $x=\frac{1}{4}$, then
$1-y^2=\frac{2}{4}\Rightarrow y^2=1-\frac{1}{2}\Rightarrow y=\pm \frac{1}{\sqrt{2}}$
$\therefore (\frac{1}{4},\pm \frac{1}{2})$ does not satisfy $1-y^2=2x$