A circle of radius 4, drawn on a chord of the parabola $y^{2} = 8 x$ as diameter, touches the axis of the parabola. Then, the slope of the chord is

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$\frac{1}{2}$

$\frac{3}{4}$

answer is C.

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Detailed Solution

Given radius of circle =4cm

Given parabola $y^{2} = 8 x$

$\Rightarrow 4 a = 8 \Rightarrow a = 2 L e t A ({at}_{1}^{2}, 2 {at}_{1}) a n d B ({at}_{2}^{2}, 2 {at}_{2}) b e e n d p o i n t s o f c h o r d f r o m t h e g i v e n d a t a c e n t r e o f c i r c l e i s m i d p o i n t o f A B \Rightarrow (\frac{a (t {_{1}}^{2} + {t_{2}}^{2})}{2}, \frac{2 a (t_{1} + t_{2})}{2}) S i n c e c i r c l e t o u c h e s a x i s a n d r a d i u s = 4 t h e n \frac{2 a (t_{1} + t_{2})}{2} = 4 \Rightarrow t_{1} + t_{2} = 2 N o w s l o p e o f A B (m) = \frac{2 a (t_{2} - t_{1})}{a (t_{2} - t_{1}) (t_{2} + t_{1})} = 1$