Parabolic question (1 Viewer)

[z600]

Find the equation of the parabola with focus S (-1,1) and the directrix y=x-2

thanks!

 

[Hikari Clover]

回复: Parabolic question

seems tricky, with directrix y=x-2 , is it really a parabola?

 

[z600]

ye equation of a parabola

 

[Hikari Clover]

回复: Re: Parabolic question

never seen a parabola with an oblique line as directrix

i need help as well:wave:

 

[wrxsti]

dont worry you wont get an oblique parabola..........

 

[Mattamz]

yeah definately wouldnt get one of these quesetions in ext.1. i seriously doubt that you get anything like this in ext.2 either.

 

[vafa]

There is no trick. It is a good question because it teaches you not to memorize mathematics. The definition of a parabola is the parabola is the locus of all points which is equidistant from a fixed point (the focus) and a fixed line (the directrix). Use this definition to solve your problem.

 

[ratpack90]

Hey Everyone, um just a quick question

The normal of the parabola x^2=18y at (-6,2) cuts the parabola again at Q. Find Q.

p.s i can do the first half just not sure how to find the coordinates of Q

I appreciate anyone's input

 

[ratpack90]

Woops sorry didn't realise this was a 3 unit thread XD

 

[SoulSearcher]

Woops sorry didn't realise this was a 3 unit thread XD

Once you get the equation of the normal to the parabola at that point, simultaneously solve the equations of the normal and the parabola.

 

[ratpack90]

ah thanks yeh can't believe i missed that lol...however when i simultaneously solved them i seemed to get the wrong answer...can you show me your working out it would be greatly appreciated

oh and thanks for the quick reply

 

[undalay]

someone tell me if i'm doing it right.

There is no trick. It is a good question because it teaches you not to memorize mathematics. The definition of a parabola is the parabola is the locus of all points which is equidistant from a fixed point (the focus) and a fixed line (the directrix). Use this definition to solve your problem.

" Find the equation of the parabola with focus S (-1,1) and the directrix y=x-2"

Let focus = A
Let the Directrix = B
Let parabola = X

AX^2 = (-1-x)^2 + (1-y)^2
AX^2 = 1+2x+x^2 + 1-2y+y^2
AX^2 = 2 +2x+x^2+2y+y^2

BX = abs ( x - y +2) / sqrt (2)
BX^2 = (x-y+2)^2 / 2
BX^2 = (x^2-2xy+4x-4y+4+y^2) / 2

Subin AX^2 = BX^2

and u get y^2 + x^2 + 8y + 2xy = 0

 

[kony]

you made a slight arithmetic mistake, but it's ok.

x² + 8x + y² + 8y + 2xy = 0
(x+4)² + (y+4)² + 2xy = 32

 

[ssglain]

you made a slight arithmetic mistake, but it's ok.
x² + 8x + y² + 8y + 2xy = 0
(x+4)² + (y+4)² + 2xy = 32

Should have looked a bit further up. ;-)

BX = abs ( x - y +2) / sqrt (2)

Simple transfer error, I presume.

BX = |x - y - 2|/√2
BX² = (x² + y² +4 - 2xy - 4x + 4y)/2

Jiggle things around and get x² + 8x + y² - 8y + 2xy = 0, which is (x + 4)² + (y - 4)² + 2xy = 32 if you prefer.

The strange thing, though, is that (x + 4)² + (y + 4)² + 2xy = 32 turns out to be a pair of parallel lines when graphed.

 

[Mattamz]

Find the equation of the parabola with axis parallel to the y-axis and passing through points (0,-2), (1,0) and (3,-8).


Thanks

let the parabola be y =ax^2 + bx + c

Sub (0,-2):
-2 = C

Sub (1,0):
0 = a + b -2
2 = a + b ____________(1)

Sub (3,-8):
-8 = 9a + 3b -2
-6 = 9a + 3b
-2 = 3a + b____________(2)

(2)-(1):
-4 = 2a
-2 = a
b = 4

therefore parabola is y = -2x^2 +4x -2

 

[Mattamz]

oh no posts are out of order...

 

[bobjingles]

Find the equation of the parabola with axis parallel to the y-axis and passing through points (0,-2), (1,0) and (3,-8).


Thanks

 

[bobjingles]

Find the equation of the parabola with axis parallel to the y-axis and passing through points (0,-2), (1,0) and (3,-8).


Thanks

 

[kittykatef]

Hey this is a bit of a general idea, but hopefully it helps.
If u find the gradient of the tangents through differentiation then find the gradient of the normal from that, then the equation of the normals (using the three points) then simultaneously solve the three equations together the result shud be the focus....:wave: jibber jabber blah blah

 

[kittykatef]

Hey my maths bro, i just had a bit of an epiphany. maybe u shud use parameters??