Find vertex , focus and directrix of conic section

**zorro** · May 3rd 2009, 10:29 PM

Find the vertex , focus and directrix of conic section
$<br /> 4y^2 + 12x - 12y +39 = 0<br />$

**stapel** · May 4th 2009, 04:22 AM

To learn the basic formulas and methods, try some online lessons.

Once you have studied at least two lessons, please attempt the exercise. If you get stuck, you can then reply with a clear listing of your work and reasoning so far. Thank you!

**zorro** · December 20th 2009, 11:52 AM

Originally Posted by stapel

To learn the basic formulas and methods, try some online lessons.

Once you have studied at least two lessons, please attempt the exercise. If you get stuck, you can then reply with a clear listing of your work and reasoning so far. Thank you!

$4y^2 + 12x -12y + 39$ = $0$

$12x - 1$ = $-4 (y^2 - 3y -10)$

$12x - 1$ = $-4 (y -5 ) (y + 2)$ .........I am stuck here !!!!

**JSB1917** · December 20th 2009, 01:30 PM

Here maybe this will help you out a little...

$4y^{2}-12y = -12x-39$

**zorro** · December 20th 2009, 02:26 PM

Originally Posted by JSB1917

Here maybe this will help you out a little...

$4y^{2}-12y = -12x-39$

$4y^{2}-12y +8= -12x-39$

$(4y-4)(y-2)= -12x-39$ .........stuck here

**HallsofIvy** · December 21st 2009, 02:43 AM

Don't factor- complete the square!

You want to arrive at $x= a(y- b)^2+ c$ . When you have that you will know that the vertex is (c, b), that the focal distance is 1/(4a), and that the directrix is given by the line x= c-1/(4a).

**JSB1917** · December 21st 2009, 09:36 AM

Originally Posted by zorro

$4y^{2}-12y +8= -12x-39$

$(4y-4)(y-2)= -12x-39$ .........stuck here

Not quite....

$4y^{2}-12y= -12x-39$

$4(y^{2}-3y)= -12x-39$

$4(y^{2}-3y+\frac{9}{4})= -12x-39+9$

$4(y-\frac{3}{2})^{2}=-12x-30$

$(y-\frac{3}{2})^{2}=-3x-\frac{15}{2}$

$(y-\frac{3}{2})^{2}=-3(x+\frac{5}{2})$

**zorro** · December 27th 2009, 10:50 AM

$\therefore y^2 = - 4ax$

$vertex = ( \frac{-5}{2} , \frac{3}{2})$

$4a = 3 => a = \frac{3}{4}$

$focus = (-a,0) = (-\frac{3}{4},0)$

directrix $x= \frac{3}{4}$ ..........Is this correct????

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Find vertex , focus and directrix of conic section

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