JEE Main 2021 — Hyperbola Question with Solution

Given,

$$x^2-2y^2=4$$

$$\Rightarrow \frac{x^2}{4}-\frac{y^2}{2}=1\Rightarrow \frac{x^2}{{(2)}^2}-\frac{y^2}{{(\sqrt{2})}^2}=1$$

Here, a = 2, $$b=\sqrt{2}$$

$$\therefore$$ $$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{2}{4}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}$$

So, Focus (F) = ($$\pm$$ a e, 0) = ($$\pm$$ $$\sqrt{6}$$, 0)

Now, equation of tangent at $$P(4,\sqrt{6})$$ is

$$x{x}_1-2y{y}_1=4$$

$$\Rightarrow x.4-2y.\sqrt{6}=4$$

$$\Rightarrow 4x-2\sqrt{6}y=4$$

$$\Rightarrow 2x-\sqrt{6}y=2$$ ....... (i)

Putting y = 0 in Eq. (i), we get x-intercept of tangent i.e. x = 1

$$\therefore$$ Q $$\equiv$$ (1, 0)

Hence, equation of corresponding latus rectum is $$x=\sqrt{6}$$

$$\therefore$$ $$R\equiv \left(\sqrt{6},\frac{2(\sqrt{6}-1)}{\sqrt{6}}\right)$$ [putting $$x=\sqrt{6}$$ in Eq. (i), we get $$y=\frac{2(\sqrt{6}-1)}{\sqrt{6}}$$]

$$\therefore$$ Area of $$\Delta QFR=\frac{1}{2}\times (QF)\times (RF)$$

$$=\frac{1}{2}(\sqrt{6}-1)\times \frac{2(\sqrt{6}-1)}{\sqrt{6}}=\frac{{(\sqrt{6}-1)}^2}{\sqrt{6}}=\left(\frac{7}{\sqrt{6}}-2\right)$$