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

1 Answer
Jan 12, 2017

#(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]}