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  • Locus of trisection point of any arbitrary double ordinate or the parabola &nb

Locus of trisection point of any arbitrary double ordinate or the parabola \mathrm{x^{2}=4by}  is

Option: 1

\mathrm{9x^{2}= by}


Option: 2

\mathrm{3x^{2}= 2by}


Option: 3

\mathrm{9x^{2}= 4by}


Option: 4

\mathrm{9x^{2}= 2by}


Answers (1)

best_answer

Let \mathrm{A \equiv\left(2 b t, b t^{2}\right), B \equiv\left(-2 b t, b t^{2}\right)}  be the extremities on the double ordinate AB.

If \mathrm{C(h, k)} be it's trisection point, then
\mathrm{3 h=4 b t-2 b t, 3 k=2 b t^{2}+b t^{2}}.

\mathrm{\Rightarrow \quad \mathrm{t}=\frac{3 \mathrm{~h}}{2 \mathrm{~b}}, \mathrm{t}^{2}=\frac{\mathrm{k}}{\mathrm{b}}}
\mathrm{\Rightarrow \quad \frac{\mathrm{k}}{\mathrm{b}}=\frac{9 \mathrm{~h}^{2}}{4 \mathrm{~b}^{2}}}

Thus locus of \mathrm{C} is \mathrm{9 x^{2}=4 b y}

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Pankaj

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