given vertex parabola at point (1,−2) and focus at (−1,−2). find the equation of parabola


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Question Number 115846 by bemath last updated on 29/Sep/20

$g i v e n v e r t e x p a r a b o l a a t p o i n t$ $(1, - 2) a n d f o c u s a t (- 1, - 2) .$ $f i n d t h e e q u a t i o n o f p a r a b o l a$
Commented by bemath last updated on 29/Sep/20

$t h a n k y o u m r P V T a n d b o b$
	Answered by 1549442205PVT last updated on 29/Sep/20

	$Since we known that in the coordinate$ $system xOy if parabol has the focus$ $F (p / 2; 0) then its equation is y^{2} = 2 px$ $and its vertex is O (0, 0) . Therefore,$ $Put X = x - 1, Y = y + 2 . Then in the$ $coordinate system XOY Parabol have$ $the focus is F (- 2, 0) with its vertex is (0, 0) .$ $Therefore, in the coordinate system$ $XOYparabol have the equation is$ $Y^{2} = - 8 X \Rightarrow - 8 (x - 1) = {(y + 2)}^{2}$ $Thus, the equation of the parabol we$ $need find is - 8 (x - 1) = {(y + 2)}^{2}$
	Answered by bobhans last updated on 29/Sep/20

	$T h e l i n e c o n n e c t i n g t h e v e r t e x a n d t h e$ $f o c u s i s t h e a x i s a n d h a s e q u a t i o n y = - 2$ $. T h e d i r e c t r i x i s t h e l i n e o r t h o g o n a l$ $t o t h e a x i s, w i t h a d i s t a n c e f r o m t h e$ $v e r t e x e q u a l t o t h e d i s t a n c e b e t w e e n$ $v e r t e x a n d f o c u s .$ $\sqrt{{(x + 1)}^{2} + {(y + 2)}^{2}} = ∣ x - 3 ∣$ ${(x + 1)}^{2} + {(y + 2)}^{2} = {(x - 3)}^{2}$ $\Leftrightarrow {(y + 2)}^{2} = {(x - 3)}^{2} - {(x + 1)}^{2}$ $\Leftrightarrow {(y + 2)}^{2} = (2 x - 2) (- 4)$ $\Leftrightarrow {(y + 2)}^{2} = - 8 (x - 1)$
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