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Q.
If the locus of a point which divides a chord with slope $2$ of the parabola $y^{2}=4 x$ internally in the ratio $1: 3$ is a parabola, then its vertex is
TS EAMCET 2019
Solution:
Let $P\left(t_{1}^{2}, 2 t_{1}\right)$ and $Q\left(t_{2}^{2}, 2 t_{2}\right)$ are extremetre of chord
$\therefore \, h=\frac{3 t_{1}^{2}+t_{2}^{2}}{4}, k=\frac{6 t_{1}+2 t_{2}}{4}$
Slope of $PQ=2$
$\therefore \, 2 =\frac{2 t_{1}-2 t_{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
$4 h=3 t_{1}^{2}+\left(1-t_{1}\right)^{2}, 4 k=6 t_{1}+2-2 t_{1}$
$4 h=4 t_{1}^{2}-2 t_{1}+1,4 k=4 t_{1}+2$
Eliminating $t_{1}$, we get
$4 h=4\left(\frac{2 k-1}{2}\right)^{2}-2\left(\frac{2 k-1}{2}\right)+1$
$4 h=4 k^{2}-6 k+3$
$\Rightarrow \left(k-\frac{3}{4}\right)^{2}=4\left(h-\frac{3}{16}\right)$
$\therefore $ Vertex $\left(\frac{3}{16}, \frac{3}{4}\right)$
