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

Option: 1

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


 


Option: 2

\mathrm{y}^2=6 \mathrm{ax}
 


Option: 3

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


Option: 4

\text{None of these}


Answers (1)

best_answer

Let \mathrm{AB} be a double ordinate, where \mathrm{\mathrm{A} \equiv\left(\mathrm{at}^2, 2 \mathrm{at}\right), \mathrm{B} \equiv\left(\mathrm{at}^2,-\mathrm{at}\right)}. If \mathrm{\mathrm{P}(\mathrm{h}, \mathrm{k})} be it's trisection point then \mathrm{ 3 \mathrm{~h}=2 \mathrm{at}^2+\mathrm{at}^2, 3 \mathrm{k}=4 \mathrm{at}-2 \mathrm{at} \Rightarrow \mathrm{t}^2=\frac{\mathrm{h}}{\mathrm{a}}, \mathrm{t}=\frac{3 \mathrm{k}}{2 \mathrm{a}}}
Thus locus is \mathrm{ \frac{9 \mathrm{k}^2}{4 \mathrm{a}^2}=\frac{\mathrm{h}}{\mathrm{a}}, i.e., 9 \mathrm{y}^2=4 \mathrm{ax}.}

Hence option 3 is correct.

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manish

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