Question

An ellipse has major \& minor axes of length $\sqrt{3}$ and 1 respectively, slides along the coordinate axes and always remains confined in the first quadrant. The locus of centre of ellipse is a circle. Find the number of tangents to the director circle of this circle which are normal to ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$.

Answer 1

Answer: 0

Let foci be $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$
and centre be $(h,k)$
$\therefore 2h=x_1+x_2$ and $2k=y_1+y_2$
here $x_1{x}_2=y_{1}y=b^2$
$\Theta S{S}^{\prime }=2ae$
$\Rightarrow {\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2=4a^2{e}^2$
$\Rightarrow {\left(x_1+x_2\right)}^2+{\left(y_1-y_2\right)}^2-4x_1{x}_2-4y_1{y}_2=4\left(a^2-b^2\right)$
$\Rightarrow 4h^2+4k^2=4\left(a^2+b^2\right)$
$\Rightarrow h^2+k^2={\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2$
$\therefore$ Locus is $x^2+y^2=1$ i.e. circle.
Normal to ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ will be
$3x\cos \theta -2y\cos \theta =5$....(i)
$\Theta (i)$ is tangent to director circle.
$\therefore$ Perpendicular distance from centre $=$ radius of director circle
$\Rightarrow \frac{5}{\sqrt{9{\sec }^2\theta +4{\mathrm{cosec}}^2\theta }}=\sqrt{2}$
$\Theta 9{\sec }^2\theta +4{\mathrm{cosec}}^2\theta =13+9{\tan }^2\theta +\frac{4}{{\tan }^2\theta }\ge 25$
$\therefore$ maximum value of $LHS=1\ne \sqrt{2}$
$\therefore$ no such $\theta$ exists.
$\therefore$ number of tangents $=0$