Search any question & find its solution
Question:
Answered & Verified by Expert
The locus of the mid-points of the focal chord of the parabola $y^{2}=4 a x$ is
Options:
Solution:
2065 Upvotes
Verified Answer
The correct answer is:
$y^{2}=2 a(x-a)$
Any chord $\mathrm{PQ}$ which bisected point $\mathrm{R}(\mathrm{h}, \mathrm{k})$ is $\mathrm{T}=\mathrm{S}$
i.e., $k y-2 a(x+h)=k^{2}-4 a h$
Since, it is a focal chord, so it must pass through focus $(a, 0)$.
$\therefore \mathrm{k}(0)-2 \mathrm{a}(\mathrm{a}+\mathrm{h})=\mathrm{k}^{2}-4 \mathrm{ah}$ $\Rightarrow \mathrm{k}^{2}=2 \mathrm{ah}-4 \mathrm{a}^{2}$ Hence, locus is
i.e., $k y-2 a(x+h)=k^{2}-4 a h$
Since, it is a focal chord, so it must pass through focus $(a, 0)$.
$\therefore \mathrm{k}(0)-2 \mathrm{a}(\mathrm{a}+\mathrm{h})=\mathrm{k}^{2}-4 \mathrm{ah}$ $\Rightarrow \mathrm{k}^{2}=2 \mathrm{ah}-4 \mathrm{a}^{2}$ Hence, locus is
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.