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

Let foci be $\left( x _1, y _1\right)$ and $\left( x _2, y _2\right)$
and centre be $(h, k)$
$\therefore 2 h = x _1+ x _2$ and $2 k = y _1+ y _2$
here $ x_1 x_2=y_1 y=b^2$
$\Theta SS ^{\prime}=2 ae$
$\Rightarrow \left( x _1- x _2\right)^2+\left( y _1- y _2\right)^2=4 a ^2 e ^2$
$\Rightarrow \left(x_1+x_2\right)^2+\left(y_1-y_2\right)^2-4 x_1 x_2-4 y_1 y_2=4\left(a^2-b^2\right)$
$\Rightarrow 4 h ^2+4 k ^2=4\left( a ^2+ b ^2\right)$
$\Rightarrow h ^2+ k ^2=\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2$
$\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
$3 x \cos \theta-2 y \cos \theta=5$....(i)
$\Theta (i)$ is tangent to director circle.
$\therefore $ Perpendicular distance from centre $=$ radius of director circle
$\Rightarrow \frac{5}{\sqrt{9 \sec ^2 \theta+4 \operatorname{cosec}^2 \theta}}=\sqrt{2}$
$\Theta 9 \sec ^2 \theta+4 \operatorname{cosec}^2 \theta=13+9 \tan ^2 \theta+\frac{4}{\tan ^2 \theta} \geq 25$
$\therefore $ maximum value of $LHS =1 \neq \sqrt{2}$
$\therefore $ no such $\theta$ exists.
$\therefore $ number of tangents $=0$