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Locus of trisection point of any double ordinate of \mathrm{y^{2}=4 a x} is

Option: 1

\mathrm{3 y^{2}=4 a x}


Option: 2

\mathrm{y}^{2}=6


Option: 3

9\mathrm{y}^{2}=4\mathrm{ax}


Option: 4

None of these


Answers (1)

best_answer

Let AB be a double ordinate, where

\mathrm{A \equiv\left(a t^{2}, 2 a t\right), B \equiv\left(a t^{2},-2 a t\right)}.
If \mathrm{P(h, k)} be it's trisection point, then

\mathrm{3 h=2 a t^{2}+a t^{2}, 3 k=4 a t-2 a t}
\mathrm{\Rightarrow \quad t^{2}=\frac{\mathrm{h}}{\mathrm{a}}, \mathrm{t}=\frac{3 \mathrm{k}}{2 \mathrm{a}}}

Thus locus is, \frac{9 \mathrm{k}^{2}}{4 \mathrm{a}^{2}}=\frac{\mathrm{h}}{\mathrm{a}}

i.e., \quad 9 \mathrm{y}^{2}=4 a x.

Posted by

Gaurav

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