find the equation of the parabola whose vertex is 2,0 and focus is -3/4,0?

Answer 1

The parabola opens to the left because the vertex is to the right side of the focus.

Let (x.y) be a point on the parabola.
The directrix is x = 2+(2+3/4) = 19/4

By definition of a parabola, the point on the parabola is equidistant from the focus and the directrix.
(x-19/4)^2 = (x+3/4)^2+y^2

Simplify,
-11(x-2) = y^2

Answer 2

Given a parabola with vertex (2,0) and focus (-3/4,0).
Find the equation of the parabola.

Since the y value for the vertex and focus are the same, the parabola opens sideways.

The equation of a parabola opening sideways is:

4p(x - h) = (y - k)²

The vertex is (h,k).
The focal length is p. That is the distance from the vertex to the focus.

So p = (-3/4) - 2 = -11/4
4p = 4(-11/4) = -11

The equation of the parabola is:

-11(x - 2) = (y - 0)²
-11(x - 2) = y²

or if you prefer
x - 2 = -y²/11
x = -y²/11 + 2

Answer 3

Vertex and focus are both on the x axis, with focus left of vertex, so parabola opens left and has equation

4p(y - k)² = (x - h), where p is focal length (here 2 3/4). so it's

11(y - 0)² = x - 2.