Question

Let $L_1{L}_1^{\prime }$ and $L_2{L}_2^{\prime }$ be the latus rectum of the ellipse $\frac{x^2}{16}+\frac{y^2}{15}=1$. If $S_1=0,S_2=0$ are the two circles having $L_1{L}_1^{\prime }$ and $L_2{L}_2^{\prime }$ as diameters, then the number of common tangents to $S_1=0$ and $S_2=0$ is

Answer 1

Given ellipse : $\frac{x^2}{16}+\frac{y^2}{15}=1$
Here,
$a^2=16,b^2=15\Rightarrow e^2=1-\frac{b^2}{a^2}\Rightarrow e=\sqrt{1-\frac{15}{16}}=\frac{1}{4}$

Endpoints of the latus rectum
$=(\pm ae,\pm \frac{b^2}{a})=(\pm 1,\pm \frac{15}{4})$

Now,
$L_1\equiv (1,\frac{15}{4}),L_1^{\prime }\equiv (1,-\frac{15}{4}),L_2\equiv (-1,\frac{15}{4}),L_2^{\prime }\equiv (-1,-\frac{15}{4})$

For $S_1=0$ center $C_1=(1,0)$ and $r_1=\frac{15}{4}$
For $S_2=0$ center $C_2=(-1,0)$ and $r_2=\frac{15}{4}$
Now,
$C_1{C}_2=2r_1+r_2=\frac{15}{2}|r_1-r_2|=0$
As $|r_1-r_2|<C_1{C}_2<r_1+r_2$
So, the circles intersect each other, therefore the number of common tangents is $2.$