Parabola Ques 3
- Through the vertex $O$ of parabola $y^2=4 x$, chords $O P$ and $O Q$ are drawn at right angles to one another. Show that for all positions of $P, P Q$ cuts the axis of the parabola at a fixed point. Also, find the locus of the middle point of $P Q$.
(1994, 4M)
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Answer:
Correct Answer: 3.$(y^2=2(x-4))$
Solution: Let the equation of chord $O P$ be $y=m x$.
Then, equation of chord will be $y=-\frac{1}{m} x$ and
$P$ is point of intersection of $y=m x$ and $y^2=4 x$ is $\left(\frac{4}{m^2}, \frac{4}{m}\right)$ and
$Q$ is point intersection of $y=-\frac{1}{m} x$ and $y^2=4 x$ is $\left(4 m^2,-4 m\right)$.
Now, equation of $P Q$ is
$ y+4 m=\frac{\frac{4}{m}+4 m}{\frac{4}{m^2}-4 m^2}\left(x-4 m^2\right) $
$\Rightarrow \quad y+4 m=\frac{m}{1-m^2}\left(x-4 m^2\right)$
$\Rightarrow \quad\left(1-m^2\right) y+4 m-4 m^3=m x-4 m^3$
$\Rightarrow \quad m x-\left(1-m^2\right) y-4 m=0$
This line meets $X$-axis, where $y=0$
i.e. $x=4 \Rightarrow O L=4$ which is constant as independent of $m$.
Again, let $(h, k)$ be the mid-point of $P Q$.
Then,
$ \begin{aligned} & h=\frac{4 m^2+\frac{4}{m^2}}{2} \\ & k=\frac{\frac{4}{m}-4 m}{2} \end{aligned} $
$\Rightarrow \quad h=2\left(m^2+\frac{1}{m^2}\right)$
and
$k=2\left(\frac{1}{m}-m\right)$
$h=2\left[\left(m-\frac{1}{m}\right)^2+2\right]$
$ k=2\left(\frac{1}{m}-m\right) $
Eliminating $m$, we get
$ 2 h=k^2+8 $
or $y^2=2(x-4)$ is required equation of locus.
BETA
