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A variable chord joining points \mathrm{P\left(a t_1{ }^2, 2 a t_1\right) \ and \ Q\left(a t_2{ }^2, 2 a t_2\right)} of the parabola \mathrm{y^2=4 a x} subtends a right angle at a fixed point\mathrm{ \left(t_0\right)} of the curve. Then PQ passes through the fixed point which is

Option: 1

\mathrm{\left(a t_0^2+4 a,-2 a t_0\right)}


Option: 2

\mathrm{\text { independent of } t_0}


Option: 3

\mathrm{\left(a t_0^2-4 a, a t_0\right)}


Option: 4

none of these


Answers (1)

best_answer

Given parabola is \mathrm{y^2=4 a x} PQ is a chord drawn at \mathrm{\left(t_1\right)} and \mathrm{\left(t_2\right)} subtends a right angle at fixed point R\mathrm{\left(t_0\right)} of the parabola y2 = 4ax such that PR ⊥ QR.

\mathrm{\therefore \quad \text { Slope of } P R \times \text { slope of } Q R=-1}

\mathrm{\therefore \quad \frac{2}{t_0+t_1} \times \frac{2}{t_0+t_2}=-1}

\mathrm{\therefore \quad-4=t_0^2+t_0\left(t_1+t_2\right)+t_1 t_2}        [i]

\mathrm{\text { The equation of chord } P Q \text { is }\left(t_1+t_2\right) y=2 x+2 a t_1 t_2}

Putting value of \mathrm{t_1 t_2} from equation (i), we get

\mathrm{\begin{aligned} & y\left(t_1+t_2\right)=2 x+2 a\left[-t_0^2-t_0\left(t_1+t_2\right)-4\right] \\ \Rightarrow & \left(y+2 a t_0\right)\left(t_1+t_2\right)=2 x-2 a t_0^2-8 a \\ \Rightarrow & \left(2 x-2 a t_0^2-8 a\right)-\left(t_1+t_2\right)\left(y+2 a t_0\right)=0 \end{aligned}} which is the

form of \mathrm{L_1+\lambda L_2=0}

Clearly, it represents a line passing through the intersection of two lines.

\mathrm{\text { Point of intersection of } L_1 \text { and } L_2 \text { is }\left(a t_0^2+4 a,-2 a t_0\right) \text {. }}

 

Posted by

Kuldeep Maurya

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