Question

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

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

Answer:

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

Step-by-step 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):

$\sqrt{(x_0-7)^2+(y_0+11)^2}$

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:

$\sqrt{(x_0-7)^2+(y_0+11)^2}=|y_0+3|$

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

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

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

$y_0=-\frac{1}{16} {x_0}^2+\frac{7}{8} x_0-\frac{161}{16}$

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:

$y=-\frac{1}{16} {x}^2+\frac{7}{8} x-\frac{161}{16}$

⇒ $y=-\frac{1}{16} (x-7)^2-7$  (in vertex form)

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