Focus and vertex of a parabola are at (3, 4) and (0,0). Find the equation of the directrix.


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Question Number 203465 by princemurtuja last updated on 19/Jan/24

$Focus and vertex of a parabola are at (3, 4) and (0, 0) .$ $Find the equation of the directrix .$
	Answered by mr W last updated on 20/Jan/24

	$F (3, 4)$ $V (0, 0)$ $d i r e c t r i x i s ⊥ t o F V .$ $e q u a t i o n f o r d i r e c t r i x :$ $3 x + 4 y + c = 0$ $e q u s t i o n f o r p a r a b o l a :$ ${(x - 3)}^{2} + {(y - 4)}^{2} = \frac{{(3 x + 4 y + c)}^{2}}{25}$ ${(0 - 3)}^{2} + {(0 - 4)}^{2} = \frac{{(3 \times 0 + 4 \times 0 + c)}^{2}}{25}$ $\Rightarrow c = 25 (- 25 r e j e c t e d)$ $d i r e c t r i x :$ $3 x + 4 y + 25 = 0$ $p a r a b o l a :$ ${(x - 3)}^{2} + {(y - 4)}^{2} = \frac{{(3 x + 4 y + 25)}^{2}}{25}$
	Commented by mr W last updated on 19/Jan/24

	Answered by som(math1967) last updated on 20/Jan/24

	$O t h e r w a y$ $i n t e r s e c t i n g p o i n t o f d i r e c t r i x$ $a n d a x i s \Rightarrow (- 3, - 4)$ $s l o p e o f a x i s = \frac{0 - 4}{0 - 3} = \frac{4}{3}$ $s l o p e o f d i r e c t r i x = \frac{- 3}{4}$ $e q u a t i o n o f d i r e c t r i x$ $(y + 4) = \frac{- 3}{4} (x + 3)$ $4 y + 16 = - 3 x - 9$ $3 x + 4 y + 25 = 0$
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