Question

if given the equation x = -1/8(y + 6)^2 + 2 what is the location of the focus point PLEASE HELP

Answer 1

Answer:

Focus is at $(0,-6)$.

Step-by-step explanation:

Given:

The equation of the parabola is: $x=-\frac{1}{8}(y+6)^{2}+2$

The standard form of a parabola in vertex form is given as:

$(y-k)^{2}=4p(x-h)$

Where, $(h,k)$ is the vertex of parabola. The value of $p$ decides the nature of parabola.

If $p >0$, the parabola opens rightwards and if $p<0$, the parabola opens leftwards.

Let us first rewrite the given equation in standard form.

$x=-\frac{1}{8}(y+6)^{2}+2\\x-2=-\frac{1}{8}(y+6)^{2}\\-8(x-2)=(y+6)^{2}\\(y-(-6))^{2}=4(-2)(x-2)$

On comparing the above equation with the standard one, we get,

$h=2,k=-6,p=-2$

Therefore, the focus of a parabola of the form $(y-k)^{2}=4p(x-h)$ is given by the point $(h+p,k)$.

∴ Focus is $(h+p,k)=(2-2,-6)=(0,-6)$