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Sagot :
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:
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 [tex](x_0,y_0)[/tex] be any point on the parabola.
Find an equation for the distance between [tex](x_0,y_0)[/tex] and the focus.
Find an equation for the distance between [tex](x_0,y_0)[/tex] and directrix. Equate these two distance equations, simplify, and the simplified equation in [tex]x_0[/tex] and [tex]y_0[/tex] is equation of the parabola.
Distance between [tex](x_0,y_0)[/tex] and the focus (7, -11):
[tex]\sqrt{(x_0-7)^2+(y_0+11)^2}[/tex]
Distance between [tex](x_0,y_0)[/tex] and the directrix, y = -3:
[tex]|y_0+3|[/tex]
Equate the two distance expressions and simplify, making [tex]y_0[/tex] the subject:
[tex]\sqrt{(x_0-7)^2+(y_0+11)^2}=|y_0+3|[/tex]
[tex](x_0-7)^2+(y_0+11)^2=(y_0+3)^2[/tex]
[tex]{x_0}^2-14x_0+49+{y_0}^2+22y_0+121={y_0}^2+6y_0+9[/tex]
[tex]{x_0}^2-14x_0+16y_0+161=0[/tex]
[tex]y_0=-\frac{1}{16} {x_0}^2+\frac{7}{8} x_0-\frac{161}{16}[/tex]
This equation in [tex](x_0,y_0)[/tex] is true for all other values on the parabola so we can rewrite with [tex](x, y)[/tex]
Therefore, the equation of the parabola with focus (7, -11) and directrix is y = -3 is:
[tex]y=-\frac{1}{16} {x}^2+\frac{7}{8} x-\frac{161}{16}[/tex]
⇒ [tex]y=-\frac{1}{16} (x-7)^2-7[/tex] (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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