y^2 = − 4px At (−(1/3),1)→ 1= −4p (− (1/3) − h) At (−(5/3),2)→ 4 = −4p (− (5/3) − h) 4 = ((− (5/3) − h)/(− (1/3) − h)) (4/3) + 4h = (5/3) + h 3h = (1/3) h = (1/9)


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Question Number 213276 by thetpainghtun_111 last updated on 02/Nov/24

$y^{2} = - 4 px$ $At (- \frac{1}{3}, 1) \to 1 = - 4 p (- \frac{1}{3} - h)$ $At (- \frac{5}{3}, 2) \to 4 = - 4 p (- \frac{5}{3} - h)$ $4 = \frac{- \frac{5}{3} - h}{- \frac{1}{3} - h}$ $\frac{4}{3} + 4 h = \frac{5}{3} + h$ $3 h = \frac{1}{3}$ $h = \frac{1}{9}$
Commented by TonyCWX08 last updated on 02/Nov/24

$W h a t i s t h i s ?$
Commented by MrGaster last updated on 02/Nov/24
My personal understanding is that this is a parabola problem.
	Answered by MrGaster last updated on 02/Nov/24

	$1 = - 4 p (- \frac{1}{3} - \frac{1}{9})$ $1 = - 4 p (- \frac{3}{9} - \frac{1}{9})$ $1 = - 4 p (- \frac{4}{p})$ $1 = 4 p \frac{4}{9}$ $9 = 16 p$ $\begin{matrix} p = \frac{9}{16} \end{matrix}$
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