Question

A parabola can be drawn given a focus of (7, -11) and a directrix of

y = -3. What can be said about the parabola?

Answer 1

Focus at(7,-11)

• x>0,y<0
• Lies in 4th quadrant

Equation of directrix y=-3

So what can be told?

• Axis of parabola=y axis

Equation of parabola

• x^2=-4ay

Answer 2

The parabola is negative, with a vertex at (7, -7) and a line of symmetry at x = 7

Explanation:

A parabola is set of all points in a plane which are an equal distance away from a given point (focus) and given line (directrix).

Let
`(x_0,y_0)` be any point on the parabola.

Find an equation for the distance between
`(x_0,y_0)` and the focus.

Find an equation for the distance between
`(x_0,y_0)` and directrix. Equate these two distance equations, simplify, and the simplified equation in
`x_0` and
`y_0` is equation of the parabola.

Distance between
`(x_0,y_0)` and the focus (7, -11):

`√((x_0-7)^2+(y_0+11)^2)`

Distance between
`(x_0,y_0)` and the directrix, y = -3:

`|y_0+3|`

Equate the two distance expressions and simplify, making
`y_0` the subject:

`√((x_0-7)^2+(y_0+11)^2)=|y_0+3|`

`(x_0-7)^2+(y_0+11)^2=(y_0+3)^2`

`{x_0}^2-14x_0+49+{y_0}^2+22y_0+121={y_0}^2+6y_0+9`

`{x_0}^2-14x_0+16y_0+161=0`

`y_0=-(1)/(16) {x_0}^2+(7)/(8) x_0-(161)/(16)`

This equation in
`(x_0,y_0)` is true for all other values on the parabola so we can rewrite with
`(x, y)`

Therefore, the equation of the parabola with focus (7, -11) and directrix is y = -3 is:

`y=-(1)/(16) {x}^2+(7)/(8) x-(161)/(16)`

⇒
`y=-(1)/(16) (x-7)^2-7` (in vertex form)

So the parabola is negative, with a vertex at (7, -7) and a vertical line of symmetry at x = 7