Respuesta :
The directrix x=-7 must be perpendicular to the axis that goes through the focus (7,0), then the axis of this parabola is y=0
The directrix cuts the axis of the parabola at the point (-7,0)
The vertex of the parabola V=(h,k) is the midpoint of the points (-7,0) and (7,0=
h=(-7+7)/2=(0)/2→h=0
k=(0+0)/2=(0)/2→k=0
Vertex: V=(h,k)→V=(0,0)
As the axis of the parabola is horizontal, the parabola has an equation of the form:
(y-k)^2=4p(x-h)
The parabola opens to the right then p is positive (p>0)
p is the distance between the vertex and the focus, the p=7
(y-k)^2=4p(x-h)
(y-0)^2=4(7)(x-0)
y^2=28x
Answer: The equation of the parabola whose focus is at (7, 0) and directrix at x = -7 is: y^2=28x
The directrix cuts the axis of the parabola at the point (-7,0)
The vertex of the parabola V=(h,k) is the midpoint of the points (-7,0) and (7,0=
h=(-7+7)/2=(0)/2→h=0
k=(0+0)/2=(0)/2→k=0
Vertex: V=(h,k)→V=(0,0)
As the axis of the parabola is horizontal, the parabola has an equation of the form:
(y-k)^2=4p(x-h)
The parabola opens to the right then p is positive (p>0)
p is the distance between the vertex and the focus, the p=7
(y-k)^2=4p(x-h)
(y-0)^2=4(7)(x-0)
y^2=28x
Answer: The equation of the parabola whose focus is at (7, 0) and directrix at x = -7 is: y^2=28x
Answer:
x = 1/28 * y^2
Step-by-step explanation:
The focal point needs to be equidistant from the directrix, which in this equation is -7 = x. With the focus being at (7, 0), at point (7, 14) as an example, it is 14 units away from the focal point, and subsequently 14 units away from the directrix, which is at -7 = x.
I don't know too much math about it, except from what I've learned in a khan academy video. I hope this helps anyone in the future.