What is the standard form for the following problem? I already know that it is a horizontal parabola. I just can't seem to be able to change it into the standard format.
$8y² +96y-12x+240 = 0$
I have gotten it to $x = 2/3 (y + 6)² - 4$, but that's not right.
Edit: How do you find the focus of this parabola?
$12x = 8(y^2+12y+36-6) = 8(y+6)^2 - 48 \to x -(-4)= \dfrac{2(y-(-6))^2}{3}$. The focus of this parabola lies on the axis of symmetry which is the line $y = -6$. We have: $F = (-4+p,-6)$, with $p = \dfrac{1}{4a} = \dfrac{1}{4\cdot \frac{2}{3}} = \dfrac{3}{8} \to F = (\frac{-29}{8},-6)$.
Edit: I think the problem might be that it is not a vertical parabola. The $y$ is squared, so it must be a sideways parabola. (Also called a horizontal parabola).
Standard form of Vertical Parabola is $(x-x_1)=4a(y-y_1)^2$,
You can easily convert your solution to this form by some rearrangement,
$(x-(-4))=4\dfrac{1}{6}(y-(-6))^2$
Where, $(x_1,y_1)$ is the vertex of the parabola that is $(-4,-6)$
You can find the focus by shifting the vertex of the parabola by $a$ on the axis of the parabola towards its opening.
