Search any question & find its solution
Question:
Answered & Verified by Expert
Through the vertex $\mathrm{O}$ of a parabola $\mathrm{y}^{2}=4 \mathrm{x}$, chords OP and $O Q$ are drawn at right angles to one another. The locus of the middle point of $\mathrm{PQ}$ is
Options:
Solution:
2122 Upvotes
Verified Answer
The correct answer is:
$y^{2}=2 x-8$
Given parabola is $\mathrm{y}^{2}=4 ...(1)
\mathrm{x}$ Let $\mathrm{P} \equiv\left(\mathrm{t}_{1}^{2}, 2 \mathrm{t}_{1}\right)$ and $\mathrm{Q} \equiv\left(\mathrm{t}_{2}^{2}, 2 \mathrm{t}_{2}\right)$
Slope of $\mathrm{OP}=\frac{2 \mathrm{t}_{1}}{\mathrm{t}_{1}^{2}}=\frac{2}{\mathrm{t}_{1}}$ and slope of $\mathrm{OQ}=\frac{2}{\mathrm{t}_{2}}$
Since $\mathrm{OP} \perp \mathrm{OQ}, \therefore \frac{4}{\mathrm{t}_{1} \mathrm{t}_{2}}=-1$ or $\mathrm{t}_{1} \mathrm{t}_{2}=-4 \ldots(2)$
Let $\mathrm{R}(\mathrm{h}, \mathrm{k})$ be the middle point of $\mathrm{PQ},$ then $\mathrm{h}=\frac{\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}}{2} \quad \ldots(3) \quad$ and $\quad \mathrm{k}=\mathrm{t}_{1}+\mathrm{t}_{2} \ldots(4)$
From $(4), \mathrm{k}^{2}=\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+2 \mathrm{t}_{1} \mathrm{t}_{2}=2 \mathrm{~h}-8 \quad[\mathrm{From}$
(2) and (3)] Hence locus of $\mathrm{R}(\mathrm{h}, \mathrm{k})$ is $\mathrm{y}^{2}=2 \mathrm{x}-8$
\mathrm{x}$ Let $\mathrm{P} \equiv\left(\mathrm{t}_{1}^{2}, 2 \mathrm{t}_{1}\right)$ and $\mathrm{Q} \equiv\left(\mathrm{t}_{2}^{2}, 2 \mathrm{t}_{2}\right)$
Slope of $\mathrm{OP}=\frac{2 \mathrm{t}_{1}}{\mathrm{t}_{1}^{2}}=\frac{2}{\mathrm{t}_{1}}$ and slope of $\mathrm{OQ}=\frac{2}{\mathrm{t}_{2}}$
Since $\mathrm{OP} \perp \mathrm{OQ}, \therefore \frac{4}{\mathrm{t}_{1} \mathrm{t}_{2}}=-1$ or $\mathrm{t}_{1} \mathrm{t}_{2}=-4 \ldots(2)$
Let $\mathrm{R}(\mathrm{h}, \mathrm{k})$ be the middle point of $\mathrm{PQ},$ then $\mathrm{h}=\frac{\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}}{2} \quad \ldots(3) \quad$ and $\quad \mathrm{k}=\mathrm{t}_{1}+\mathrm{t}_{2} \ldots(4)$
From $(4), \mathrm{k}^{2}=\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+2 \mathrm{t}_{1} \mathrm{t}_{2}=2 \mathrm{~h}-8 \quad[\mathrm{From}$
(2) and (3)] Hence locus of $\mathrm{R}(\mathrm{h}, \mathrm{k})$ is $\mathrm{y}^{2}=2 \mathrm{x}-8$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.