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  • A variable chord PQ of the parabola  subtends a right angle at the vertex.

A variable chord PQ of the parabola \mathrm{y=4 x^2 } subtends a right angle at the vertex. Equation of the locus of the points of intersection of the normals at P and Q is represented by\mathrm{ 8 y=8 x^2+p}, then p is

Option: 1

3


Option: 2

4


Option: 3

5


Option: 4

6


Answers (1)

best_answer

\mathrm{\text { Parametric point on the parabola } y=4 x^2 \text { is }\left(t, 4 t^2\right)}

\mathrm{\text { Let } P \equiv\left(t_1, 4 t_1^2\right) \text { and } Q \equiv\left(t_2, 4 t_2^2\right)}

If A is vertex of the parabola then

\mathrm{\text { Slope of } A P \times \text { Slope of } A Q=-1 \quad(\because A P \perp A Q)}

\mathrm{\therefore 4 t_1 \times 4 t_2=-1 \Rightarrow t_1 t_2=-\frac{1}{16}}

Equations of normals P and Q  respectively, are 

\mathrm{x+8 t_1 y=t_1+32 t_1^3}           ...[1]

\mathrm{\text { and } x+8 t_2 y=t_2+32 t_2^3}      ...[2]

Let the normals (1) and (2) intersect at (h, k) then solving (1) and (2), we get

\mathrm{\begin{aligned} & k= \frac{1+32\left(t_1^2+t_2^2+t_1 t_2\right)}{8} \\ & \Rightarrow \quad k=\frac{1+32\left(t_1^2+t_2^2\right)+32 t_1 t_2}{8} \quad=\frac{1+32\left(\left(t_1+t_2\right)^2-2 t_1 t_2\right)-2}{8} \quad\left(\because 16 t_1 t_2=-1\right) \\ &=4\left\{\left(t_1+t_2\right)^2+\frac{1}{8}\right\}-\frac{1}{8} \Rightarrow k=4\left(t_1+t_2\right)^2+\frac{3}{8} \quad \ldots(3) \end{aligned}}

\mathrm{\text { and } h=-32 t_1 t_2\left(t_1+t_2\right) \Rightarrow h=2\left(t_1+t_2\right)}       ....[4]

\mathrm{\text { Eliminating } t_1 \text { and } t_2 \text { from ( } 3 \text { ) and (4) we get }}

\mathrm{k=h^2+\frac{3}{8} \Rightarrow 8 k=8 h^2+3}

\mathrm{\text { Hence locus of }(h, k) \text { is } 8 y=8 x^2+3}

\mathrm{\text { But equation of locus is } 8 y=8 x^2+p}

\mathrm{\therefore p=3}

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manish

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