15:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 16:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 17:T474,

Focii of hyperbola $$( \pm a e, 0)=( \pm 2,0)$$

$$\Rightarrow \quad a e=2$$ ..... (i)

Let equation of hyperbola be $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

It passes through $$P(\sqrt{2}, \sqrt{3})$$

$$\begin{aligned} & \Rightarrow \frac{2}{a^2}-\frac{3}{b^2}=1 \Rightarrow \frac{2}{a^2}-\frac{3}{a^2\left(e^2-1\right)}=1 \\ & \Rightarrow \quad \frac{2}{a^2}-\frac{3}{a^2 e^2-a^2}=1 \Rightarrow \frac{2}{a^2}-\frac{3}{4-a^2}=1 \end{aligned}$$

Let $$a^2=x$$

$$\Rightarrow \quad \frac{2}{x}-\frac{3}{4-x}=1$$

$$\Rightarrow \quad 8-2 x-3 x=4 x-x^2$$

$$\Rightarrow \quad x^2-9 x+8=0$$

$$\Rightarrow \quad(x-1)(x-8)=0$$

$$\Rightarrow \quad x=1,8$$

$$\Rightarrow \quad {a}^2=1,8$$

$$\Rightarrow \quad b^2=4-a^2=3,-4$$

But, $$b^2 \neq-4$$

$$\therefore \quad a^2=1, b^2=3$$

So, equation of hyperbola is $$\frac{x^2}{1}-\frac{y^2}{3}=1$$

Equation of tangent at $$P(\sqrt{2}, \sqrt{3})$$ is

$$\begin{aligned} & \frac{x \cdot \sqrt{2}}{1}-\frac{y \cdot \sqrt{3}}{3}=1 \\ & \Rightarrow \quad y=x \sqrt{6}-\sqrt{3} \end{aligned}$$

Hyperbola

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