Discussion Forum

 If the line x-y=-4K is a tangent  to the parabola $y^{2}=8x$ at P, then the perpendicular distance of normal at P from (K,2K) is 

Answer:

Explanation:

Since, x-y=-4k  or y=x+4k is a tangent to the parabola $y^{2}=8x$, therefore

    $4k=\frac{2}{1}$            $[\because  4a=8 \Rightarrow a=2]$

  $\Rightarrow$       $k=\frac{1}{2}$

 Also  points of concept  p is    $\left(\frac{a}{m^{2}},\frac{2a}{m}\right)=(2,4)$

  Now, equation of normal to the $y^{2}=8x$ at (2,4) is 

$(y-4)=\frac{-4}{2(2)}(x-2)$

 [$\because$   Equation of normal to the parabola , $y^{2}=4ax$ at

                                        $(x_{1},y_{1})is y-y_{1}=\frac{-b}{2a}(x-x_{1})$

$\Rightarrow$   $y-4=-1(x-2)$

$\Rightarrow$   $y-4=-x+2$

$\Rightarrow$   $x+y=6$

$\Rightarrow$   $x+y-6=0$

The perpendicular distance of normal from (k,2k) ie  $(\frac{1}{2},1)$

$=\frac{|\frac{1}{2}+1-6|}{\sqrt{1^{2}+1^{2}}}=\frac{|\frac{3}{2}-6|}{\sqrt{2}}=\frac{9}{2\sqrt{2}}$