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 $
$\Rightarrow y_{1}^{2}=1-2 x_{1}$