Let E be an ellipse whose axes are parallel to the co-ordinates...

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

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

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

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)$</p><p>\left(h + a, k\right) = \left(5, - 4\right)$

\left(h + ae, k\right) = \left(4, - 4\right)$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/0323cc0c-5292-493b-894c-5a34c1360e5e-image.png" style="width: 282px; height: 164px;" /></p><p>Clearly,$a = 2 \text{ and } ae = 1$</p><p>$⇒ e = \dfrac{1}{2}$</p><p>We know that,${b}^{2} = {a}^{2}\left(1 - {e}^{2}\right)$</p><p>${b}^{2} = {a}^{2} - {a}^{2}{e}^{2}$</p><p>${b}^{2} = {2}^{2} - 1$</p><p>$⇒ {b}^{2} = 3$</p><p>So, equation of ellipse$\dfrac{{\left(x - 3\right)}^{2}}{4} + \dfrac{{\left(y + 4\right)}^{2}}{3} = 1$</p><p>Intersecting with given tangent</p><p>$\dfrac{{x}^{2} - 6x + 9}{4} + \dfrac{{m}^{2}{x}^{2}}{3} = 1$</p><p>$⇒ 3\left({x}^{2} - 6x + 9\right) + 4{m}^{2}{x}^{2} = 12$</p><p>$⇒ 3{x}^{2} - 18x + 27 + 4{m}^{2}{x}^{2} = 12$</p><p>$⇒ \left(3 + 4{m}^{2}\right){x}^{2} - 18x + 15 = 0$</p><p>Now,$D = 0$(as it is tangent)</p><p>$⇒ {\left(18\right)}^{2} = 4\left(3 + 4{m}^{2}\right) \times 15$</p><p>So,$5{m}^{2} = 3 .