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  • If a chord PQ makes a right angle at vertex of parabola and <img alt="\mathrm

If a chord PQ makes a right angle at vertex of parabola \mathrm{y^2=4 a x} and \mathrm{R} be the mid-point of \mathrm{ P Q} such that abscissa of \mathrm{P, R, Q } are in G.P., hence find the coordinates \mathrm{P} and \mathrm{Q.}

Option: 1

\mathrm{(a, \pm 2 a)}


Option: 2

\mathrm{\left(\frac{a}{4}, \pm a\right)}


Option: 3

\mathrm{(4 a, \pm 4 a)}


Option: 4

\mathrm{\text { None }}


Answers (1)

best_answer

Let  \mathrm{P\left(\left(a t_1^2, 2 a t_1\right)\right.} and \mathrm{Q\left(a t_2^2, 2 a t_2\right) \Rightarrow} abscissa of R is \mathrm{\frac{a\left(t_1^2+t_2^2\right)}{2}}

\mathrm{\Rightarrow \quad \mathrm{at}_1^2, \frac{\mathrm{a}\left(\mathrm{t}_1^2+\mathrm{t}_2^2\right)}{2}, \mathrm{at}_2^2 \text { are in G.P. and } \mathrm{t}_1 \mathrm{t}_2=-4}

\mathrm{ \begin{aligned} & \frac{\left(t_1^2+t_2^2\right)^2}{4}=t_1^2 \cdot t_2^2=16 \Rightarrow t_1^2+t_2^2=8 \\ \Rightarrow & \left(t_1+t_2\right)^2=8-8=0 \Rightarrow t_1=-t_2 \\ \Rightarrow & t_1^2=4 \Rightarrow t_1= \pm 2 . \end{aligned} }
Hence the points P and Q are \mathrm{(4 a, \pm 4 a).}

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