Question

How do you find an equation of the parabola with vertex (3,2) and focus (1,2)?

Answer 1

`(y-2)^2=-8(x-3)` that expands to
`y^2+8x-4y-20=0`

Explanation:

Focus S and Vertex V:

(1, 2)S-------V(1, 2)

The level SV is y = the common value of `y_V and y_S= 2.`

V is ahead of S. So, the axis is in the negative

x-direction.

The perpendicular tangent at the vertex VT is given by `x = x_V=3`.

Now, the equation of this parabola, with size a = 2, axis `y = 2 larr`

and tangent at the vertex x = 2 is

`(y-2)^2=-4(2)(x-3)`, that expands to

`y^2+8x-4y-20=0`

graph{((y-2)^2+8(x-3))(x-3)(y-2)=0 [-20, 20, -10, 10]}