Question

Consider a circle $C:x^2+y^2-8y+12=0$ 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 $\left(a^2-2b^2\right)$.

Answer 1

Answer: 8

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