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

Answer 1

Answer: (A), (C)

Point $P$ is $(\alpha ,\beta )$
Let Point $E$ is $\left(x_1,y_1\right)$
given circle is $x^2+y^2=1...(1)$
$\therefore {\alpha }^2+{\beta }^2=1$
Also tangent at $P(\alpha ,\beta )$ is
$\alpha x+\beta y=1$
Put $(1,y)$
$\alpha +\beta y_1=1...(2)$
Now Slope of $OP=$ Slope of $OE$
$\frac{\beta }{\alpha }=\frac{y_1}{x_1}$
$\alpha =\frac{x_1\beta }{y_1}...(3)$
Put in (2)
$\beta =\frac{y_1}{x_1+y_1^2}$
$\therefore \alpha =\frac{x_1}{x_1+y_1^2}$
Put $\alpha ,\beta$ in $(1)$ we get
$\frac{x_1^2}{{\left(x_1+y_1^2\right)}^2}+\frac{y_1^2}{{\left(x_1+y_1^2\right)}^2}=1$
$\Rightarrow y_1^2=1-2x_1$