Question

If the tangent to ellipse $x^2+2y=1$ at point $P(\frac{1}{\sqrt{2}},\frac{1}{2})$ meets the auxiliary circle at the points $R$ and $Q$, then tangents to circle at $Q$ and $R$ intersect at

• A. $(\frac{1}{\sqrt{2}},1)$
• B. $(1,\frac{1}{\sqrt{2}})$
• C. $(\frac{1}{2},\frac{1}{2})$
• D. $(\frac{1}{2},\frac{1}{\sqrt{2}})$

Answer 1

The correct option is A $(\frac{1}{\sqrt{2}},1)$

Equation of tangent to ellipse at given point is
$x\left(\frac{1}{\sqrt{2}}\right)+2y\left(\frac{1}{2}\right)=1$
$\Rightarrow x+\sqrt{2}y=\sqrt{2}$ ...(i)
Now, $QR$ is chord of contact of circle $x^2+y^2=1$ with respect to point $T(h,K)$.
Then, $QR\equiv hx+Ky=1$ ...(ii)
Equations (i) and (ii) represent same straight line, then comparing ratio of coefficients, we have
$\frac{h}{1}=\frac{K}{\sqrt{2}}=\frac{1}{\sqrt{2}}$
Hence, $(h,K)\equiv (\frac{1}{\sqrt{2}},1)$