Mathematics - Ellipse Question with Solution | TestHub

The given equation can be written as $\frac{x^2}{{\alpha }^2}+\frac{y^2}{4}=1$

Any point on the curve is $\mathrm{P}(\alpha \cos \theta ,2\sin \theta )$

$\therefore \mathrm{PQ}=\sqrt{{\alpha }^2{\cos }^2\theta +4(1+\sin \theta {)}^2}$

Let $\mathrm{S}(\theta )={\alpha }^2{\cos }^2\theta +4(1+\sin \theta {)}^2=\left(4+{\alpha }^2\right)+8\sin \theta +\left(4-{\alpha }^2\right){\sin }^2\theta$

Now, $\frac{dS}{d\theta }=2\cos \theta \left[4+\left(4-{\alpha }^2\right)\sin \theta \right]$

$=\frac{\mathrm{ds}}{\mathrm{d}\theta }=0$ gives $\theta =\frac{\pi }{2},\frac{3\pi }{2}$ or $\sin \theta =\frac{4}{{\alpha }^2-4}$.

But $0<{\alpha }^2<8$ and $|\sin \theta |\le 1$, so $\sin \theta =\frac{4}{{\alpha }^2-4}$ is not possible.

Also, at $\theta =\frac{3\pi }{2},P$ becomes $(0,-2)$ which is same as point $Q$.

So $PQ=0$.

Also, ${\frac{{\mathrm{d}}^2\mathrm{S}}{\mathrm{d}{\theta }^2}]}_{\theta =\frac{\pi }{2}}=-8\sin \theta +2\left(4-{\alpha }^2\right)\cos 2\theta$

$=-8+2\left(4-{\mathrm{a}}^2\right)(-1)$ $=-2\left(8-{\alpha }^2\right)<0($ As $0<{\alpha }^2<8)$

Hence $\theta =\frac{\pi }{2}$ corresponds to the maximum values of $S$.

So, $P$ is $(0,2)$ Clearly, reflection of $\mathrm{P}(0,2)$ in the x -axis is $\mathrm{R}(0,-2)$.

So, least distance of $R(0,-2)$ from the line $3x-4y+7=0$, is $\frac{|3(0)-4(-2)+7|}{\sqrt{(3{)}^2+(-4{)}^2}}=\frac{15}{5}=3$