Answer:
The standard form of the equation of the parabola will be [tex]y=\frac{x^2}{32}[/tex].
Step-by-step explanation:
To find he equation of the parabola with a focus at (0, 8) and
a directrix at y = -8, we may use the distance formula.
[tex]\sqrt{\left(x-0\right)^2+\left(y-8\right)^2}=\sqrt{\left(x-x\right)^2+\left(y+8\right)^2}[/tex]
[tex]\mathrm{Square\:both\:sides}[/tex]
[tex]\left(\sqrt{\left(x-0\right)^2+\left(y-8\right)^2}\right)^2=\left(\sqrt{\left(x-x\right)^2+\left(y+8\right)^2}\right)^2[/tex] [tex]......A[/tex]
Solving
[tex]\left(\sqrt{\left(x-0\right)^2+\left(y-8\right)^2}\right)^2[/tex]
[tex]\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}=a^{\frac{1}{2}}[/tex]
[tex]=\left(\left(\left(x-0\right)^2+\left(y-8\right)^2\right)^{\frac{1}{2}}\right)^2[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \left(a^b\right)^c=a^{bc}[/tex]
[tex]=\left(\left(x-0\right)^2+\left(y-8\right)^2\right)^{\frac{1}{2}\cdot \:2}[/tex]
[tex]=\left(x-0\right)^2+\left(y-8\right)^2[/tex] ∵ [tex]\frac{1}{2}\cdot \:2=1[/tex]
[tex]=x^2+y^2-16y+64[/tex] ∵ [tex]\mathrm{Expand\:}\left(x-0\right)^2+\left(y-8\right)^2:\quad x^2+y^2-16y+64[/tex]
Similarly
[tex]\mathrm{Expand\:}\left(\sqrt{\left(x-x\right)^2+\left(y+8\right)^2}\right)^2:\quad \left(y+8\right)^2[/tex]
So, the equation A becomes
[tex]x^2+y^2-16y+64=\left(y+8\right)^2[/tex]
[tex]x^2+y^2-16y+64=y^2+16y+64[/tex]
[tex]y^2-16y=y^2+16y-x^2[/tex]
[tex]-32y=-x^2[/tex]
[tex]\frac{-32y}{-32}=\frac{-x^2}{-32}[/tex]
[tex]y=\frac{x^2}{32}[/tex]
Therefor, the standard form of the equation of the parabola will be [tex]y=\frac{x^2}{32}[/tex].
