Question

Derive the equation of the parabola with a focus at (0, −4) and a directrix of y = 4. f(x) = −16x2 f(x) = 16x2 f(x) = − x2 f(x) = x2

Answer 1

The equation of the parabola is y = -1/16 x²

Explanation:

* Lets revise some facts about the parabola

- Standard form equation for a parabola of vertex at (0 , 0)

- If the equation is in the form x² = 4py, then

- The axis of symmetry is the y-axis, x = 0

- 4p equal to the coefficient of y in the given equation to

solve for p

- If p > 0, the parabola opens up.

- If p < 0, the parabola opens down.

- Use p to find the coordinates of the focus, (0 , p)

- Use p to find equation of the directrix , y = - p

* Lets solve the problem

∵ The focus at (0 , -4)

∵ The coordinates of the focus are (0 , p)

∴ p = -4

∵ The directrix is y = 4

∵ The equation of the directrix , y = - p

∴ -p = 4 ⇒ p = -4

∵ the equation is in the form x² = 4py

∵ p = -4

∴ x² = 4(-4)y

∴ x² = -16y ⇒ divide both sides by -16

∴ y = -1/16 x²

* The equation of the parabola is y = -1/16 x²

Answer 2

`f(x)=-(1)/(16) x^2`

Explanation:

We are to derive the equation of the parabola with a focus at
`(0, -4)` and a directrix of y = 4.

We know that a parabola is the locus of all the points as long as the distance from the fixed point on the parabola to the fixed line directrix is kept same.

This parabola is facing downwards. So assuming any point on the parabola to be
`(x,y)`.

Distance from focus
`(0,-4)` to
`(x,y)` =
`\sqrt{(x-0)^(2) +(y+4) ^(2)}=\sqrt{x^(2)+ (y+4)^(2)}`

Distance from
`(x,y)` to directrix
`(y=4)` =
`\left | y-4 \right |`

Equating these distances as they are equal:

`\sqrt{x^(2)+ (y+4)^(2)}=\left | y-4 \right |`

`{x^(2)+ (y+4)^(2)=(y-4)^(2)`

`x^2+y^2+8 y +16 = y^2 - 8 y+16`

`x^2 = -8 y - 8 y= -16 y`

`x^2= - 16 y`

So the equation of the parabola is
`f(x)=-(1)/(16) x^2`.