Question

Let $E$ be an ellipse whose axes are parallel to the co-ordinates axes, having its center at $(3,-4)$, one focus at $(4,-4)$ and one vertex at $(5,-4)$. If $mx-y=4,m>0$ is a tangent to the ellipse $E$, then the value of $5m^2$ is equal to

Answer 1

Given: center $C(3,-4)$, focus $F_1(4,-4)$, vertex $V_1(5,-4)$
$\because ae=C{F}_1=1$ and $a=C{V}_1=2$ so, $b=\sqrt{3}$
$E:\frac{(x-3{)}^2}{4}+\frac{(y+4{)}^2}{3}=1$
Equation of tangent
$y+4=m(x-3)\pm \sqrt{4m^2+3}\Rightarrow y=mx-3m-4\pm \sqrt{4m^2+3}$
Comparing with $y=mx-4$
We get $-3m\pm \sqrt{4m^2+3}=0$
$\Rightarrow 9m^2=4m^2+3$
$\Rightarrow 5m^2=3$