Find the vertex , focus and directrix of conic section
$\displaystyle
4y^2 + 12x - 12y +39 = 0
$
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!
Not quite....
$\displaystyle 4y^{2}-12y= -12x-39$
$\displaystyle 4(y^{2}-3y)= -12x-39$
$\displaystyle 4(y^{2}-3y+\frac{9}{4})= -12x-39+9$
$\displaystyle 4(y-\frac{3}{2})^{2}=-12x-30$
$\displaystyle (y-\frac{3}{2})^{2}=-3x-\frac{15}{2}$
$\displaystyle (y-\frac{3}{2})^{2}=-3(x+\frac{5}{2})$
$\displaystyle \therefore y^2 = - 4ax$
$\displaystyle vertex = ( \frac{-5}{2} , \frac{3}{2})$
$\displaystyle 4a = 3 => a = \frac{3}{4}$
$\displaystyle focus = (-a,0) = (-\frac{3}{4},0)$
directrix $\displaystyle x= \frac{3}{4}$..........Is this correct????