Question

Consider a circle $C : x ^2+ y ^2-8 y +12=0$ and an ellipse $E : \frac{ x ^2}{ a ^2}+\frac{ y ^2}{ b ^2}=1( a > b$ and $b < 2)$. If the maximum perpendicular distance from the foci of the ellipse upon the tangent drawn to the circle is 7 units, and shortest distance between both the curves is 1 unit, then find the value of $\left(a^2-2 b^2\right)$.

Answer 1

$C: x^2+(y-4)^2=4$
$E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $
$\sqrt{a^2 e^2+16}+2=7 $
$\Rightarrow a^2 e^2=9 \Rightarrow a e=3 $
$\text { and } 2-b=1 \Rightarrow b=1 $
$b^2=a^2\left(1-e^2\right) \Rightarrow 1=a^2-9 \Rightarrow a^2=10$
$\therefore \frac{x^2}{10}+\frac{y^2}{1}=1 $
$a^2-2 b^2=10-2=8 $