Question

The number of point on the ellipse $\frac{x^2}{50}+\frac{y^2}{20}=1$ from which pair of perpendicular tangents are drawn to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is

• A. $2$
• B. $1$
• C. $0$
• D. $4$

Answer 1

The correct option is D $4$

We have two ellipses,
$\frac{x^2}{50}+\frac{y^2}{20}=1\cdots \cdots \left(1\right)$
$\frac{x^2}{16}+\frac{y^2}{9}=1\cdots \cdots \left(2\right)$

Comparing with the standard equation of ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

In ellipse $\left(1\right):$
$a_1=\sqrt{50},b_1=\sqrt{20}$
$\Rightarrow a_1=5\sqrt{2},b_1=2\sqrt{5}$

And in ellipse $\left(2\right)$:
$a_2=4,b_2=3$

Plotting them on the coordinate plane.

We can clearly observe, ellipse $\left(2\right)\text{is inside ellipse}\left(1\right).$
$\frac{x^2}{16}+\frac{y^2}{9}=1,$ perpendicular tangents should come from its director circle.

Director circle is given by,
$x^2+y^2=a_2^2+b_2^2$
$\Rightarrow x^2+y^2=16+9=25$
Thus, the radius of the director circle is $5$ units.
Director circle of ellipse $\left(2\right)$ will pass through $(0,\pm 5)$ and $(\pm 5,0)$

Now the interesting aspect is that these four points which are on the director circle,are on this bigger ellipse too. So it seems as if the perpendicular tangents are drawn from this ellipse

$\text{Hence, required number of points}=4.$