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The normal at $\mathrm{(a, 2 a)}$ meet the curve $\mathrm{ y^2=4 a x}$ again at $\mathrm{ \left(a t^2, 2 a t\right)}$ then the value of $\mathrm{ t}$ equals

Option: 1
$1$

Option: 2
$-1$

Option: 3
$-3$

Option: 4
$3$

Answers (1)

best_answer

If the normal at $\mathrm{\left(a t_1{ }^2, 2 a t_1\right)}$ w.r.t. parabola $\mathrm{y^2=4 a x}$ meet the curve again at $\mathrm{\left(a t_2{ }^2, 2 a t_2\right)}$ then we have
$\mathrm{ t_2=-t_1-\frac{2}{t_1} }$ ........(i)
Here, we have $\mathrm{a t_1{ }^2=a}$ and

$\mathrm{ 2 a t_1=2 a \Rightarrow t_1=1 }$

$\mathrm{ For \: t_2 \: putting \: t_1=1\: in \: (i), we\: get }$

$\mathrm{ t_2=-1-\frac{2}{1}=-3 }$

Hence option 3 is correct.

Posted by

manish painkra

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