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Question Number 90989 by ajfour last updated on 27/Apr/20

Commented by ajfour last updated on 27/Apr/20

$I f e q . o f p a r a b o l a i s 9 y = 4 (x^{2} - 9)$ $a n d A B = B C = C D, f i n d e q . o f$ $t h e l i n e A D .$
	Answered by mr W last updated on 27/Apr/20

	$s a y e q n . o f p a r a b o l a i s g e n e r a l l y$ $y = {p x}^{2} - q$ $A (a, 0)$ $D (0, - d)$ $e q n . o f A D :$ $\frac{x}{a} - \frac{y}{d} = 1 \Rightarrow y = d (\frac{x}{a} - 1)$ $i n t e r s e c t i o n o f l i n e w i t h p a r a b o l a :$ ${p x}^{2} - q = d (\frac{x}{a} - 1)$ ${a p x}^{2} - d x + a (d - q) = 0$ $x_{1} + x_{2} = \frac{d}{a p}$ $x_{1} x_{2} = \frac{d - q}{p}$ $w e h a v e$ $\frac{x_{1} + x_{2}}{2} = \frac{a}{2} \Rightarrow \frac{d}{a p} = a \Rightarrow d = {p a}^{2} . . . (i)$ $x_{1} - x_{2} = \frac{a}{3}$ ${(x_{1} - x_{2})}^{2} = {(x_{1} + x_{2})}^{2} - 4 x_{1} x_{2} = \frac{a^{2}}{9}$ $\Rightarrow a^{2} - \frac{4 (d - q)}{p} = \frac{a^{2}}{9}$ $\Rightarrow \frac{2 a^{2}}{9} = \frac{d - q}{p}$ $\Rightarrow 2 {p a}^{2} = 9 d - 9 q . . . (i i)$ $7 {p a}^{2} = 9 q$ $\Rightarrow a = 3 \sqrt{\frac{q}{7 p}}$ $\Rightarrow d = \frac{9 q}{7}$ $\Rightarrow e q n . y = 3 \sqrt{\frac{p q}{7}} x - \frac{9 q}{7}$
	Commented by ajfour last updated on 27/Apr/20

	$T h a n k s S i r, t o o g o o d a n d g r e a t,$ $y o u s o l v e d i t g e n e r a l l y!$
	Commented by mr W last updated on 27/Apr/20

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