Parabola Ques 2
- Show that the locus of a point that divides a chord of slope $2$ of the parabola $y^2=4 a x$ internally in the ratio $1: 2$ is a parabola. Find the vertex of this parabola.
(1995, 5M)
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Answer:
Correct Answer: 2.$\left(\frac{2}{9}, \frac{8}{9}\right)$
Solution: Let $A\left(t_1^2, 2 t_1\right)$ and $B\left(t_2^2, 2 t_2\right)$ be coordinates of the end points of a chord of the parabola $y^2=4 x$ having slope $2$ .
Now, slope of $A B$ is
$ m=\frac{2 t_2-2 t_1}{t_2^2-t_1^2}=\frac{2\left(t_2-t_1\right)}{\left(t_2-t_1\right)\left(t_2+t_1\right)}=\frac{2}{t_2+t_1} $
But $m=2 \quad $ [given]
$\Rightarrow \quad 2=\frac{2}{t_2+t_1}$
$\Rightarrow \quad t_1+t_2=1$ $\quad$ ……..(i)
Let $P(h, k)$ be a point on $A B$ such that, it divides $A B$ internally in the ratio $1: 2$.
Then,
$ h=\frac{2 t_1^2+t_2^2}{2+1} \text { and } h=\frac{2\left(2 t_1\right)+2 t_2}{2+1} $
$ \Rightarrow \quad 3 h=2 t_1^2+t_2^2 $ $\quad$ ……..(ii)
and $ \quad 3 h=4 t_1+ 2t_2$ $\quad$ ……..(iii)
On substituting value of $t_1$ from Eq. (i) in Eq. (iii)
$ 3 k=4\left(1-t_2\right)+2 t_2 $
$ \Rightarrow \quad 3 k=4-2 t_2 $
$ \Rightarrow \quad t_2=2-\frac{3 k}{2}$ $\quad$ ……..(iv)
On substituting $t_1=1-t_2$ in Eq. (ii), we get
$ \begin{aligned} & 3 h=2\left(1-t_2\right)^2+t_2^2 \\ & = \quad 2\left(1-2 t_2+t_2^2\right)+t_2^2 \\ & = \quad 3 t_2^2-4 t_2+2=3\left(t_2^2-\frac{4}{3} t_2+\frac{2}{3}\right) \\ & = \quad 3\left[\left(t_2-\frac{2}{3}\right)^2+\frac{2}{3}-\frac{4}{9}\right]=3\left(t_2-\frac{2}{3}\right)^2+\frac{2}{3} \\ & \Rightarrow \quad 3 h-\frac{2}{3}=3\left(t_2-\frac{2}{3}\right)^2 \\ & \Rightarrow \quad 3\left(h-\frac{2}{9}\right)=3\left(2-\frac{3 k}{2}-\frac{2}{3}\right)^2 \quad \text { [from Eq. (iv)] } \\ & \Rightarrow \quad 3\left(h-\frac{2}{9}\right)=3\left(\frac{4}{3}-\frac{3 k}{2}\right)^2 \\ & \Rightarrow \quad\left(h-\frac{2}{9}\right)=\frac{9}{4}\left(k-\frac{8}{9}\right)^2 \\ & \Rightarrow \quad\left(h-\frac{8}{9}\right)^2=\frac{4}{9}\left(h-\frac{2}{9}\right) \\ \end{aligned} $
On generalising, we get the required locus
$ \left(y-\frac{8}{9}\right)^2=\frac{4}{9}\left(x-\frac{2}{9}\right) $
This represents a parabola with vertex at $\left(\frac{2}{9}, \frac{8}{9}\right)$.
BETA
