First slide

Parabola

Question

$Let the normal to parabola y^{2} = 4 a x at P meets the curve again in Q . If P Q and the$ $normal at Q makes angles α and β respectively with the positive x -axis in positive$ $direction, then \tan α (\tan α + \tan β) is equal to$

Moderate

Solution

$Let P = (a t_{1}^{2}, 2 a t_{1}) & Q = (a t_{2}^{2}, 2 a t_{2})$
$Equation of the normal at P (t_{1}) is y + x t_{1} = 2 a t_{1} + a t_{1}^{3} \dots \dots \dots \dots . (1)$
$Slope of the normal at P (t_{1}) is - t_{1} = \tan α$
$Similarly Slope of the normal at Q (t_{2}) is - t_{2} = \tan β$
$(1) is passing through Q (a t_{2}^{2}, 2 a t_{2})$
$\begin{array}{l} \Rightarrow 2 a t_{2} + a t_{2}^{2} t_{1} = 2 a t_{1} + a t_{1}^{3} \\ \Rightarrow 2 a (t_{2} - t_{1}) = a t_{1} (t_{1}^{2} - t_{2}^{2}) \\ \Rightarrow - 2 a (t_{1} - t_{2}) = a t_{1} (t_{1} + t_{2}) (t_{1} - t_{2}) \\ \Rightarrow - 2 = t_{1} (t_{1} + t_{2}) \\ \Rightarrow - 2 = - \tan α (- \tan α - \tan β) \\ \Rightarrow \tan α (\tan α + \tan β) = - 2 \end{array}$

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