JEE Main 2021 — Ellipse Question with Solution

We know that, If the standard ellipse is shifted to have center $\left(h, k\right)$, then it's equation is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

The center is $C\left(h, k\right)$
The vertices on the major axis are $A\left(h+a, k\right)$ and $A^{\text{'}}\left(h-a, k\right)$
The foci are $S\left(h+ae, k\right)$ and $S^{\text{'}}\left(h-ae, k\right)$

The eccentricity $e=\sqrt{\frac{a^2-b^2}{a^2}}$

Given $C\left(3,-4\right),S\left(4,-4\right)$ and $A\left(5,-4\right)$

Therefore, $\left(h, k\right)=\left(3, -4\right)$

$\left(h+a, k\right)=\left(5, -4\right)$

$\left(h+ae, k\right)=\left(4, -4\right)$

Clearly, $a=2 \&ae=1$

$\Rightarrow e=\frac{1}{2}$

We know that, $b^2=a^2\left(1-e^2\right)$

$b^2=a^2-a^2{e}^2$

$b^2=2^2-1$

$\Rightarrow b^2=3$

So, equation of ellipse $\frac{{\left(x-3\right)}^2}{4}+\frac{{\left(y+4\right)}^2}{3}=1$

Intersecting with given tangent

$\frac{x^2-6x+9}{4}+\frac{m^2{x}^2}{3}=1$

$\Rightarrow 3\left(x^2-6x+9\right)+4m^2{x}^2=12$

$\Rightarrow 3x^2-18x+27+4m^2{x}^2=12$

$\Rightarrow \left(3+4m^2\right)x^2-18x+15=0$

Now, $D=0$ (as it is tangent)

$\Rightarrow {\left(18\right)}^2=4\left(3+4m^2\right)\times 15$

So, $5m^2=3.$