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Zepler [3.9K]
2 years ago
15

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

Mathematics
2 answers:
tigry1 [53]2 years ago
7 0

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

shtirl [24]2 years ago
3 0

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

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