find-the-equation-of-a-parabola-with-focus-3-3-and-directrix-y-0-

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

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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