JEE Main 2022 — Hyperbola Question with Solution

The equation of tangent to hyperbola $$x^2-y^2=1$$ within slope $$m$$ is equal to $$y=mx\pm \sqrt{m^2-1}$$ ...... (i)

And for same slope $$m$$, equation of tangent to ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ is $$y=mx\pm \sqrt{a^2{m}^2+b^2}$$ ...... (ii)

$$\because$$ Equation (i) and (ii) are identical

$$\therefore$$ $$a^2{m}^2+b^2=m^2-1$$

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

But equation of common tangent is $$y=\sqrt{\frac{5}{2}}x+k$$

$$\therefore$$ $$m=\sqrt{\frac{5}{2}}\Rightarrow \frac{5}{2}=\frac{1+b^2}{1-a^2}$$

$$\therefore$$ $$5a^2+2b^2=3$$ ....... (i)

eccentricity of ellipse $$=\frac{1}{\sqrt{2}}$$

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

$$\Rightarrow a^2=2b^2$$ ....... (ii)

From equation (i) and (ii) : $$a^2=\frac{1}{2},b^2=\frac{1}{4}$$

$$\therefore$$ $$4(a^2+b^2)=3$$