find the equation of a parabola with focus (3,3) and directrix y = 0


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Question Number 88378 by Rio Michael last updated on 10/Apr/20

$find the equation of a parabola with focus (3, 3)$ $and directrix y = 0$
Commented by john santu last updated on 10/Apr/20

$o y e s . s h o u l d b e a v e r t e x$ $a t (3, \frac{3}{2})$
Commented by john santu last updated on 10/Apr/20

Commented by john santu last updated on 10/Apr/20
$focus (a, b+p) = (3,3) ⇒ { ((a=3)),((b+p = 3)) :} directrix y = b−p = 0 , b=p b = p = (3/2) equation of a parabola (x−3)^2 = 4.((3/2))(y−(3/2)) (x−3)^2 = 6 (y−(3/2))= 6y−9$
$f o c u s (a, b + p) = (3, 3) \Rightarrow {\begin{cases} a = 3 \\ b + p = 3 \end{cases}$ $d i r e c t r i x y = b - p = 0, b = p$ $b = p = \frac{3}{2}$ $e q u a t i o n o f a p a r a b o l a$ ${(x - 3)}^{2} = 4 . (\frac{3}{2}) (y - \frac{3}{2})$ ${(x - 3)}^{2} = 6 (y - \frac{3}{2}) = 6 y - 9$
Commented by mr W last updated on 10/Apr/20

$c o r r e c t$
	Answered by mr W last updated on 10/Apr/20

	$d e f i n i t i o n o f p a r a b o l a :$ ${(x - 3)}^{2} + {(y - 3)}^{2} = {(y - 0)}^{2}$ $x^{2} - 6 x - 6 y + 18 = 0$ $\Rightarrow y = \frac{x^{2}}{6} - x + 3$
	Commented by jagoll last updated on 10/Apr/20

	$if the focus at (2, 5) and dirextrix$ $x + 4 = 0,$ $the equation should be$ ${(x - 2)}^{2} + {(y - 5)}^{2} = {(x + 4)}^{2}$ $- 4 x + 4 + y^{2} - 10 y + 25 = 8 x + 16$ $y^{2} - 10 y + 13 = 12 x$ $y^{2} - 12 x - 10 y + 13 = 0 ?$ $it correct ?$
	Commented by mr W last updated on 10/Apr/20

	$y e s .$ $b u t w h a t i f t h e f o c u s i s a t (2, 5) a n d t h e$ $d i r e x t r i x i s y + 2 x + 1 = 0 ?$
	Commented by jagoll last updated on 10/Apr/20

	$does this matter wrong sir ?$
	Commented by mr W last updated on 10/Apr/20

	$y o u r a n s w e r i s c o r r e c t .$ $i a s k e d a n o t h e r q u e s t i o n : w h e n t h e$ $d i r e x t r i x i s y + 2 x + 1 = 0, w h a t i s t h e n$ $t h e e q n . o f t h e p a r a b o l a ?$
	Commented by john santu last updated on 10/Apr/20

	$i f f o c u s a t (2, 5) a n d d i r e x t r i x i s$ $y + 2 x + 1 = 0, w e g e t p = \frac{1}{2} ∣ \frac{5 + 4 + 1}{\sqrt{5}} ∣$ $p = \sqrt{5}, t h e l i n e p a s s e s t h r o u g h$ $f o c u s o r t h o g o n a l t o d i r e x t r i x$ $\Rightarrow x - 2 y + 6 = 0$ $t h e p a r a b o l a 4 p (x - x_{v}) = {(y - y_{v})}^{2}$ $b y r o t a t e d θ = \tan^{- 1} (\frac{1}{2}) . i t c o r r e c t$
	Commented by som(math1967) last updated on 10/Apr/20

	${(x - 2)}^{2} + {(y - 5)}^{2} = \frac{{(2 x + y + 1)}^{2}}{2^{2} + 1^{2}}$
	Commented by mr W last updated on 10/Apr/20

	$f o c u s a t (2, 5)$ $d i r e x t r i x 2 x + y + 1 = 0$ $e q n . o f p a r a b o l a a c c . t o d e f i n i t i o n :$ ${(x - 2)}^{2} + {(y - 5)}^{2} = {(\frac{2 x + y + 1}{\sqrt{2^{2} + 1^{2}}})}^{2}$ $\Rightarrow x^{2} + 4 y^{2} - 4 x y - 24 x - 52 y + 144 = 0$
	Commented by mr W last updated on 10/Apr/20

	Commented by Rio Michael last updated on 10/Apr/20

	$wow am so grateful sirs$
	Commented by john santu last updated on 10/Apr/20

	$h o w i f t h e v e r t e x p a r a b o l a i s$ $(2, - 1) a n d d i r e x t r i x y - 2 x + 2 = 0 ?$
	Commented by jagoll last updated on 10/Apr/20

	$great sir$
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