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

##### 1 Answer
Jan 12, 2017

${\left(y - 2\right)}^{2} = - 8 \left(x - 3\right)$ that expands to
${y}^{2} + 8 x - 4 y - 20 = 0$

#### Explanation:

Focus S and Vertex V:

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

The level SV is y = the common value of ${y}_{V} \mathmr{and} {y}_{S} = 2.$

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 $y = 2 \leftarrow$

and tangent at the vertex x = 2 is

${\left(y - 2\right)}^{2} = - 4 \left(2\right) \left(x - 3\right)$, that expands to

${y}^{2} + 8 x - 4 y - 20 = 0$

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