Question

Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8.

y = one divided by thirty twox2
y2 = 8x
y2 = 32x
y = one divided by eightx2

Answer 1

The standard form of the equation of the parabola will be
`y=(x^2)/(32)`.

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.

`√(\left(x-0\right)^2+\left(y-8\right)^2)=√(\left(x-x\right)^2+\left(y+8\right)^2)`

`\mathrm{Square\:both\:sides}`

`\left(√(\left(x-0\right)^2+\left(y-8\right)^2)\right)^2=\left(√(\left(x-x\right)^2+\left(y+8\right)^2)\right)^2`
`......A`

Solving

`\left(√(\left(x-0\right)^2+\left(y-8\right)^2)\right)^2`

`\mathrm{Apply\:radical\:rule}:\quad √(a)=a^{(1)/(2)}`

`=\left(\left(\left(x-0\right)^2+\left(y-8\right)^2\right)^{(1)/(2)}\right)^2`

`\mathrm{Apply\:exponent\:rule}:\quad \left(a^b\right)^c=a^(bc)`

`=\left(\left(x-0\right)^2+\left(y-8\right)^2\right)^{(1)/(2)\cdot \:2}`

`=\left(x-0\right)^2+\left(y-8\right)^2` ∵
`(1)/(2)\cdot \:2=1`

`=x^2+y^2-16y+64` ∵
`\mathrm{Expand\:}\left(x-0\right)^2+\left(y-8\right)^2:\quad x^2+y^2-16y+64`

Similarly

`\mathrm{Expand\:}\left(√(\left(x-x\right)^2+\left(y+8\right)^2)\right)^2:\quad \left(y+8\right)^2`

So, the equation A becomes

`x^2+y^2-16y+64=\left(y+8\right)^2`

`x^2+y^2-16y+64=y^2+16y+64`

`y^2-16y=y^2+16y-x^2`

`-32y=-x^2`

`(-32y)/(-32)=(-x^2)/(-32)`

`y=(x^2)/(32)`

Therefor, the standard form of the equation of the parabola will be
`y=(x^2)/(32)`.