For what point of the parabola $y^{2} = 4 a x$ is (1) the normal equal to twice the sub-tangent, (2) the normal equal to the difference between the sub-tangent and the sub-normal ?

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Solution

For $y^{2} = 4 a x$ at $(a t^{2}, 2 a t)$
Length of sub tangent $P Q = 2 a t^{2}$

Length of subnormal $M N = 2 a$

Length of normal $P N = 2 a \sqrt{1 + t^{2}}$

(1) Length of sub tangent $=$ length of normal

$2 a t^{2} = 2 a \sqrt{1 + t^{2}} t^{2} = \sqrt{1 + t^{2}} t^{4} = 1 + t^{2} t^{4} - t^{2} - 1 = 0$

using quadratic formula

$t^{2} = \frac{1 \pm \sqrt{1 - 4 (1) (- 1)}}{2} t^{2} = \frac{1 \pm \sqrt{5}}{2} \Rightarrow t^{2} = \frac{1 + \sqrt{5}}{2} t = \sqrt{\frac{1 + \sqrt{5}}{2}}$

Point of contact is $(a t^{2}, 2 a t)$

$\Rightarrow ⎛ ⎝ a \frac{1 + \sqrt{5}}{2}, 2 a \sqrt{\frac{1 + \sqrt{5}}{2}} ⎞ ⎠ \Rightarrow (\frac{a + a \sqrt{5}}{2}, a \sqrt{2 + 2 \sqrt{5}})$

(2) normal $=$ difference between sub tangent and su normal

$y^{2} = 4 a x (a t^{2}, 2 a t)$
Given : $2 a \sqrt{1 + t^{2}} = 2 a t^{2} - 2 a$
$\sqrt{1 + t^{2}} = t^{2} - 1 1 + t^{2} = t^{4} + 1 - 2 t^{2} t^{4} - 3 t^{2} = 0 t^{2} (t^{2} - 3) = 0 t^{2} - 3 = 0 t^{2} = 3 \Rightarrow t = \sqrt{3}$

Point of contact is $(a t^{2}, 2 a t)$

$\Rightarrow (3 a, 2 \sqrt{3} a)$

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