prove that the locus of middle point of the normal chord of the parabola y^2 =4ax is (y^2 /(2a))+((4a^3 )/y^2 )=x−2a


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Question Number 47454 by peter frank last updated on 10/Nov/18

$prove that the locus$ $of middle point of the$ $normal chord of the$ $parabola y^{2} = 4 ax is$ $\frac{y^{2}}{2 a} + \frac{{4 a}^{3}}{y^{2}} = x - 2 a$
	Answered by ajfour last updated on 10/Nov/18

	$M (h, k)$ $y = 2 a t; x = {a t}^{2}$ $M i s m i d p o i n t o f c h o r d, s o$ $\Rightarrow k - 2 {a t}_{1} = - t_{1} (h - {a t}_{1}^{2}) . . . . (i)$ $2 a (t_{2} + t_{1}) = 2 k . . . (i i)$ $a (t_{2}^{2} + t_{1}^{2}) = 2 h . . . (i i i)$ $\Rightarrow \frac{2 a (t_{2} - t_{1})}{a (t_{2}^{2} - t_{1}^{2})} = - t_{1}$ $\Rightarrow t_{1} (t_{1} + t_{2}) = - 2 . . . (i v)$ $u s i n g (i i) i n (i v)$ $t_{1} (\frac{k}{a}) = - 2$ $\Rightarrow t_{1} = \frac{- 2 a}{k}$ $u s i n g t h i s i n (i)$ $k - 2 a (\frac{- 2 a}{k}) = (\frac{2 a}{k}) [h - a (\frac{4 a^{2}}{k^{2}})]$ $N o w (h, k) \to (x, y)$ $(y + \frac{4 a^{2}}{y}) (\frac{y}{2 a}) = x - \frac{4 a^{3}}{y^{2}}$ $\Rightarrow \frac{y^{2}}{2 a} + 2 a = x - \frac{4 a^{3}}{y^{2}}$ $o r \frac{y^{2}}{2 a} + \frac{4 a^{3}}{y^{2}} = x - 2 a .$
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