Discussion Forum

If the locus of a point which divides a chord with slope 2 of the parabola $y^{2}=4x$ , internally in the ratio 1:3 is a parabola , then its vertes is 

Answer:

Explanation:

 Let   $P( t_{1}^{2},2t_{1})$    and    $Q( t_{2}^{2},2t_{2})$  are extremetre of chord

$\therefore$        $h=\frac{3 t_{1}^{2}+t_{2}^{2}}{4},k=\frac{6t_{1}+2t_{2}}{4}$

.Slope of PQ=2

 $\therefore$    $2=\frac{2t_{1}-2t_{2}}{t_{1}^{2}-t_{2}^{2}}$

 $t_{1}+t_{2}=1 $  $\Rightarrow  t_{2}=1-t_{1}$

 Put the value of $t_{2}$ in h and k , we get

 $4h=3 t_{1}^{2}+(1-t_{1})^{2}, 4k=6t_{1}+2-2t_{1}$

 $4h=4t_{1}^{2}-2t_{1}+1, 4k=4t_{1}+2$

 Eliminating $t_{1}$ , we get

$4h=4\left(\frac{2k-1}{2}\right)^{2}-2\left(\frac{2k-1}{2}\right)+1$

 $4h=4k^{2}-6k+3$

 $\Rightarrow$     $\left(k-\frac{3}{4}\right)^{2}=4\left(h-\frac{3}{16}\right)$

 $\therefore$    Vertex  ($\frac{3}{16},\frac{3}{4})$