Discussion Forum

The area (in sq units)  of the quadrilateral formed by the tangent at the endpoints of the latus rectum to  the ellipse $\frac{x^{2}}{5}+\frac{y^{2}}{5}=1$   is 

Answer:

Explanation:

 Given equation of ellipse is 

   $\frac{x^{2}}{5}+\frac{y^{2}}{5}=1$  

  $\therefore$     $a^{2}=9,b^{2}=5$  

$\Rightarrow$     a=3,  $b=\sqrt{5}$

Now,    $e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{5}{9}}=\frac{2}{3}$

   foci    $=  (\pm ae, 0)= (\pm 2,0)$    and    $\frac{b^{2}}{a}=\frac{5}{3}$

$\therefore$   Extremities of one latusrectum are

  $(2,\frac{5}{3})$

   $(2,\frac{-5}{3})$

 $\therefore$   Equation  of tangent at    $(2,\frac{5}{3})$  is

             $\frac{x(2)}{9}+\frac{y(5/3)}{5}=1$

  or 2x+3y= 9.......(ii)

   Eq .(ii)  intersects X and Y axes at   $(\frac{9}{2},0)$

     and (0,3) , respectively

  $\therefore$   Area of quadrilateral

                     =   $4\times$  Area of   $\triangle POQ$

                      =  $4\times \left(\frac{1}{2} \times \frac{9}{2}\times 3\right)=27  sq.unit$