Discussion Forum

Given A circle , $2x^{2}+2y^{2}=5$  and a parabola , $y^{2}=4\sqrt{5}x$

  Statement I , An equation of common tangne to these curves is $y=x+\sqrt{5}$

 Statement II, If the line  $y=mx+\frac{\sqrt{5}}{m}(m\neq0)$   is the common tangent, then m satisfies   $m^{4}-3m^{2}+2=0$.

Answer:

Explanation:

Equation of circle can be rewritten as 

$x^{2}+y^{2}=\frac{5}{2}$,

    Centre → (0,) and radius   $\rightarrow\sqrt{\frac{5}{2}}$

 Let common tangent be

                        $y=mx+\frac{\sqrt{5}}{m}$

  $\Rightarrow m^{2}x-my+\sqrt{5}=0$

 The perependicular from centre to the tangent is equal to radius of the cirtcle

$\therefore$                         $\frac{\frac{\sqrt{5}}{m}}{\sqrt{1+m^{2}}}=\sqrt{\frac{5}{2}}$

             $\Rightarrow  $     $ m\sqrt{1+m^{2}}=\sqrt{2}$

  $\Rightarrow$     $   m^{2}(1+m^{2})=2$

  $\Rightarrow $    $  m^{4}+m^{2}-2=0$

   $\Rightarrow $    $  (m^{2}+2)(m^{2}-1)=0$

$\Rightarrow $    $  m\pm1$

                                      ($\because m^{2}+2\neq0,$  as   $m\in R)$

   $\therefore$      $y=\pm(x+\sqrt{5})$  both statements are correct as m=$\pm$ 1 satisfies the given equation of statement II.