Question

If tangent at a variable point $P$ on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ intersects the ellipse $\frac{x^2}{15}+\frac{y^2}{10}=1$ at the points $A$ and $B$, then the locus of the points of intersection of the tangents at $A$ and $B$ is

• A. an ellipse with eccentricity $\frac{1}{2}$
• B. an ellipse with eccentricity $\frac{1}{\sqrt{2}}$
• C. a circle with radius $5$ units
• D. a circle with radius $\sqrt{5}$ units

Answer 1

The correct option is C a circle with radius $5$ units


Let equation of the tangent at $P\left(\theta \right)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ be
$\frac{x\cos \theta }{3}+\frac{y\sin \theta }{2}=1\cdots \left(1\right)$

Now $AB$ is the chord of contact of $Q(h,k)$ with respect to $\frac{x^2}{15}+\frac{y^2}{10}=1$
$\therefore$ Equation of $AB$ is
$T=0\Rightarrow \frac{hx}{15}+\frac{ky}{10}=1\cdots \left(2\right)$

$\left(1\right)$ and $\left(2\right)$ represent the same line, so
$\frac{\left(\frac{h}{15}\right)}{\left(\frac{\cos \theta }{3}\right)}=\frac{\left(\frac{k}{10}\right)}{\left(\frac{\sin \theta }{2}\right)}=1\Rightarrow \cos \theta =\frac{h}{5},\sin \theta =\frac{k}{5}\Rightarrow h^2+k^2=25$

$\therefore$ The locus of $Q$ is a circle with radius $5$ units.