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Parametric Equation of Normal

The normal a ...

Question

The normal a point $(b t_{1}^{2}, 2 b t_{1})$ on a parabola meets the parabola again in the point $(b t_{2}^{2}, 2 b t_{2})$ then

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Solution

The correct option is B
Let $P = (a t_{1}^{2}, 2 a t_{1})$ , $Q = (a t_{2}^{2} .2 a t_{2})$ Since the normals at P and Q meet on the parabola, $t_{1} t_{2} = 2$
Point of intersection of the normals $(x_{1} y_{1}) = (a [(t_{1}^{2} + t_{2}^{2} + t_{1} t_{2} + 2], - a t_{1} t_{2} [t_{1} + t_{2}])$
$\Rightarrow x_{1} = a (t_{1}^{2} + t_{2}^{2} + t_{1} t_{2} + 2) = a (t_{1}^{2} t_{2}^{2} + 2) \Rightarrow a (t_{1}^{2} + t_{2}^{2}) x_{1} - 4 a$
$P Q^{2} = (t_{1}^{2} + t_{2}^{2})^{2} + (2 a t_{1} - 2 a t_{2})^{2} = a^{2} (t_{1} - t_{2})^{2} [(t_{1} + t_{2})^{2} + 4] = a (t_{1}^{2} + t_{2}^{2} - 4) a (t_{1}^{2} + t_{2}^{2} + 8) = (x_{1} - 8 a) (X_{1} + 4 a)$

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