Discussion Forum

If a hyperbola passes through the point $P(\sqrt{2},\sqrt{3})$ and has foci at (± 2,0), then the tangent to this hyperbola at P also passes through the point

Answer:

Explanation:

Let the equation of hyperbola be

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

$\therefore$       $ae=2\Rightarrow a^{2}e^{2}=4$

$\Rightarrow    a^{2}+b^{2}=4$

$\Rightarrow    b^{2}=4-a^{2}$

 $\therefore$   $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{4-a^{2}}=1$

  Since ($\sqrt{2},\sqrt{3}$) lie on hyperbola

$\therefore$    $\frac{2}{a^{2}}-\frac{3}{4-a^{2}}=1$

     $\Rightarrow$     $8-2a^{2}-3a^{2}=a^{2}(4-a^{2})$

$\Rightarrow$   $8-5a^{2}=4a^{2}-a^{4}$

$\Rightarrow$     $a^{4}-9a^{2}+8=0$

$\Rightarrow$     $(a^{4}-8)(a^{4}-1)=0$

$\Rightarrow$     $a^{4}=8,a^{4}=1$

$\therefore$     a=1

   Now equation of hyperbola is 

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

$\therefore$  Equation of tangent at ($\sqrt{2},\sqrt{3}$) is given by

          $\sqrt{2}x-\frac{\sqrt{3}y}{3}=1$

  $\Rightarrow$    $\sqrt{2}x-\frac{y}{\sqrt{3}}=1$

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