JEE Main 2022 — Ellipse Question with Solution

Given ellipse $$x^2+a^2{y}^2=25a^2\Rightarrow \frac{x^2}{25a^2}+\frac{y^2}{25}=1$$

eccentricity $$(e_1)=\sqrt{1-\frac{b^2}{a^2}}$$

$$=\sqrt{1-\frac{25}{25a^2}}$$

$$=\sqrt{1-\frac{1}{a^2}}$$

$$\Rightarrow e_1^2=1-\frac{1}{a^2}$$

Also, given hyperbola,

$$x^2-a^2{y}^2=5$$

$$\Rightarrow \frac{x^2}{5}-\frac{y^2}{\frac{5}{a^2}}=1$$

eccentricity $$(e_2)=\sqrt{1+\frac{b^2}{a^2}}$$

$$=\sqrt{1+\frac{5}{5a^2}}$$

$$=\sqrt{1+\frac{1}{a^2}}$$

$$\Rightarrow e_2^2=1+\frac{1}{a^2}$$

Also given,

$$e_1=b\times e_2$$

$$\Rightarrow e_1^2=b^2\times e_2^2$$

$$\Rightarrow 1-\frac{1}{a^2}=b^2\left(1+\frac{1}{a^2}\right)$$

$$\Rightarrow \frac{a^2-1}{a^2}=\frac{b^2(a^2+1)}{a^2}$$

$$\Rightarrow b^2=\frac{a^2-1}{a^2+1}$$

$$y={\log }_e^x$$ is inverse of $$y=e^x$$ so it is mirror image of each other with respect to y = x line.

Slope of tangent to y = e^x curve

$$\frac{dy}{dx}=e^x$$

Slope of tangent to $$y={\log }_e^x$$ curve,

$$\frac{dy}{dx}=\frac{1}{x}$$

Both tangents are parallel to y = x line for minimum distance condition.

$$\therefore$$ Slope of y = x line = Slope of both the tangent.

$$\therefore$$ $$\frac{dy}{dx}=e^x=1\Rightarrow e^x=e^0=x=0$$

$$\therefore$$ $$y=e^x=e^0=1$$

and $$\frac{dy}{dx}=\frac{1}{x}=1\Rightarrow x=1$$

$$\therefore$$ $$y={\log }_e^1=0$$

$$\therefore$$ tangent at (0, 1) point of $$y=e^x$$ curve and tangent at (1, 0) point of $$y={\log }_e^x$$ curve are parallel.

$$\therefore$$ Minimum distance between point (0, 1) and (1, 0) is $$=\sqrt{1^2+1^2}=\sqrt{2}$$

$$\therefore$$ $$a=\sqrt{2}$$

So, $$b^2=\frac{a^2-1}{a^2+1}=\frac{2-1}{2+1}=\frac{1}{3}$$

$$\therefore$$ $$a^2+\frac{1}{b^2}=2+\frac{1}{1/3}=5$$