Answer:
Focus is at [tex](0,-6)[/tex].
Step-by-step explanation:
Given:
The equation of the parabola is: [tex]x=-\frac{1}{8}(y+6)^{2}+2[/tex]
The standard form of a parabola in vertex form is given as:
[tex](y-k)^{2}=4p(x-h)[/tex]
Where, [tex](h,k)[/tex] is the vertex of parabola. The value of [tex]p[/tex] decides the nature of parabola.
If [tex]p >0[/tex], the parabola opens rightwards and if [tex]p<0[/tex], the parabola opens leftwards.
Let us first rewrite the given equation in standard form.
[tex]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)[/tex]
On comparing the above equation with the standard one, we get,
[tex]h=2,k=-6,p=-2[/tex]
Therefore, the focus of a parabola of the form [tex](y-k)^{2}=4p(x-h)[/tex] is given by the point [tex](h+p,k)[/tex].
∴ Focus is [tex](h+p,k)=(2-2,-6)=(0,-6)[/tex]
