Question

A hyperbola passes through the point $P(\sqrt{2},\sqrt{3})$and has foci at $(\pm 2,0)$.

Then the tangent to this hyperbola at $P$ also passes through the point

• A. $(2\sqrt{2},3\sqrt{3})$
• B. $(\sqrt{3},\sqrt{2})$
• C. $(-\sqrt{2},-\sqrt{3})$
• D. $(3\sqrt{2},2\sqrt{3})$

Answer 1

The correct option is A

$(2\sqrt{2},3\sqrt{3})$

Explanation for the correct option:

Step 1. The general equation of a hyperbola is

$\frac{x^2}{a^2}–\frac{y^2}{b^2}=1$

Given, A hyperbola passes through the point $P(\sqrt{2},\sqrt{3})$and has foci at $(\pm 2,0)$.

Hence,

$\begin{array}{rcl}{a}^2+b^2& =& 4\end{array}$ and

$\begin{array}{rcl}\frac{2}{a^2}–\frac{3}{b^2}& =& 1\end{array}$

Step 2: Finding the variables for the equation

$\begin{array}{rcl}\left(\frac{2}{4–b^2}\right)–\left(\frac{3}{b^2}\right)& =& 1\end{array}$

$\Rightarrow$ $\begin{array}{rcl}2b^2-3(4-b^2)& =& b^2(4-b^2)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}2b^2-12+3b^2& =& 4b^2-b^4\end{array}$

$\Rightarrow$$\begin{array}{rcl}{b}^4-4b^2+2b^2+3b^2-12& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{b}^4+b^2-12& =& 0\end{array}$

$\begin{array}{rcl}{b}^2& =& \frac{-1\pm \sqrt{1-4(-12)}}{2}\\ & =& \frac{-1\pm \sqrt{49}}{2}\\ & =& \frac{-1\pm 7}{2}\\ & =& \frac{-8}{2},\frac{6}{2}\\ & =& -4,3\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{b}^2& =& 3\end{array}$

and $\begin{array}{rcl}{a}^2& =& 1\end{array}$

Hence, $x^2-\frac{y^2}{3}=1$

Step 3: Finding the point that the tangent passes through

The tangent at the point $P(\sqrt{2},\sqrt{3})$is$\sqrt{2}x–\frac{y}{\sqrt{3}}=1$

Clearly, it passes through the point $(2\sqrt{2},3\sqrt{3})$

$\therefore$the tangent to the hyperbola at the point $P$ also passes through the point $(2\sqrt{2},3\sqrt{3})$

Hence, option (A) is correct,