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The normals at \mathrm{P, Q} the ends of a focal chord of a parabola meets the parabola again in \mathrm{\mathrm{P}^{\prime}} and \mathrm{Q^{\prime}} respectively. \mathrm{P Q} is

parallel to \mathrm{P^{\prime} Q^{\prime}}and that \mathrm{P Q=k P^{\prime} Q^{\prime}}, where \mathrm{k=}

Option: 1

\frac{1}{2}


Option: 2

\frac{1}{3}


Option: 3

2


Option: 4

3


Answers (1)

best_answer

Let \mathrm{P\left(a t_1^2, 2 a t_1\right)} and \mathrm{Q\left(a t_2^2, 2 a t_2\right)} be the ends of a focal chord of the parabola \mathrm{y^2=4 a x. }

\mathrm{\therefore \mathrm{t}_1 \mathrm{t}_2=-1,}

Let \mathrm{P^{\prime} \equiv\left(a^{t_3^2}, 2 a t_3\right) \, \, and \, \, Q^{\prime} \equiv\left(a^{t_4^2}, 2 a t_4\right)}
Then, \mathrm{t_3=-t_1-\frac{2}{t_1}=-t_1+2 t_2\, \, and\, \, t_4=-t_2+2 t_1}

Slope, \mathrm{m_{P Q}=\frac{2 a t_2-2 a t_1}{a t_2^2-a t_1^2}=\frac{2}{t_1+t_2}}

     \mathrm{ \begin{aligned} & \mathrm{m}_{P^{\prime} Q^{\prime}}=\frac{2}{t_3+t_4}=\frac{2}{t_1+t_2}=m_{P Q} \\\\ & \therefore P Q \| P^{\prime} Q^{\prime} \end{aligned} }

 

Now, \mathrm{P Q=\sqrt{\left(a t_2^2-a t_1^2\right)^2+\left(2 a t_2-2 a t_1\right)^2}}

\mathrm{=a\left(t_2-t_1\right) \sqrt{\left(t_2+t_1\right)^2+4}, taking\, \, t_2>t_1.}

Similarly \mathrm{P^{\prime} Q^{\prime}=\sqrt{a^2\left(t_4-t_3\right)^2\left\{\left(t_4+t_3\right)+4\right\}}}

\mathrm{=\sqrt{\mathrm{a}^2\left(3 \mathrm{t}_1-3 \mathrm{t}_2\right)^2\left\{\left(\mathrm{t}_1+\mathrm{t}_2\right)^2+4\right\}}=3 \mathrm{a}\left(\mathrm{t}_2-\mathrm{t}_1\right) \sqrt{\left(\mathrm{t}_2+\mathrm{t}_1\right)^2+4}}

Hence, \mathrm{3 P Q=P^{\prime} Q^{\prime}}

Posted by

Gaurav

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