Question:
Let $\mathrm{E}$ 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 $m x-y=4, m>0$ is a tangent to the ellipse $E$, then the value of $5 \mathrm{~m}^{2}$ is equal to
Solution:
Given $C(3,-4), S(4,-4)$
and $\mathrm{A}(5,-4)$
Hence, $a=2 \& a e=1$
$\Rightarrow \mathrm{e}=\frac{1}{2}$
$\Rightarrow \mathrm{b}^{2}=3$
So, $\mathrm{E}: \frac{(\mathrm{x}-3)^{2}}{4}+\frac{(\mathrm{y}+4)^{2}}{3}=1$
Intersecting with given tangent.
$\frac{x^{2}-6 x+9}{4}+\frac{m^{2} x^{2}}{3}=1$
Now, $D=0$ (as it is tangent)
So, $5 \mathrm{~m}^{2}=3$
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