# Thread: Find vertex , focus and directrix of conic section

1. ## Find vertex , focus and directrix of conic section

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

2. 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!

3. ## I am stuck

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 !!!!

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

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

5. ## I tried this but still no go

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

6. 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).

7. Originally Posted by zorro
$4y^{2}-12y +8= -12x-39$

$(4y-4)(y-2)= -12x-39$.........stuck here
Not quite....

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

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

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

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

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

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

8. $\therefore y^2 = - 4ax$

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

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

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

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

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### Find the vertex,focus and the directrix of the parabola 4y2 12x-12y 39=0.

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