Q69 / 120 · VITEEE 2011 · Mathematics

The locus of the mid-points of the focal chord of the parabola $y^{2}=4 a x$ is

  • A
    $y^{2}=a(x-a)$
  • Correct answer
    B
    $y^{2}=2 a(x-a)$
  • C
    $y^{2}=4 a(x-a)$
  • D
    None of these

Solution

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

Hence, the correct answer is option: 'B'.

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