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10-2. Parabola, Ellipse, Hyperbola
hard
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 $\mathrm{E}$, then the value of $5 \mathrm{~m}^{2}$ is equal to $.....$
A
$1$
B
$2$
C
$3$
D
$4$
(JEE MAIN-2021)
Solution
and $\mathrm{A}(5,-4)$
Hence, $\mathrm{a}=2\, \,\mathrm{ae}=1$
$\Rightarrow \mathrm{e}=\frac{1}{2}$
$\Rightarrow \mathrm{b}^{2}=3$
So, $E: \frac{(x-3)^{2}}{4}+\frac{(y+4)^{2}}{3}=1$
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, $5 \mathrm{~m}^{2}=3$.
Standard 11
Mathematics
