# Question #75d40

Jan 4, 2018

Focus of the parabola is $\left(6 , - 1\right)$

${x}^{2} = 16 y$

#### Explanation:

Given -

$12 \left(x - 3\right) = {\left(y + 1\right)}^{2}$

It can be written as -

${\left(y + 1\right)}^{2} = 12 \left(x - 3\right)$

Since it is in the form ${y}^{2} = 4 a x$; it opens to the right

This is in the form-

${\left(y - k\right)}^{2} = 4 a \left(x - h\right)$

Where -

$\left(h , k\right)$ is vertex
$a$ is distance of focus or directrix from the vertex.

From the given equation we can find the value of $a$. For this, we have to rewrite the equation like this.

${\left(y + 1\right)}^{2} = 4 \times 3 {\left(x - 3\right)}^{3}$

Then $a = 3$
Vertex $\left(3 , - 1\right)$

focus $\left(6 , - 1\right)$ PROBLEM 2

Given -

Focus $\left(0 , 4\right)$
Directrix $y = - 4$

Look at the graph Vertex is a point of the parabola.

Vertex lies at equi distance in between focus and directrix.

The coordinates of the point at which the directrix cuts the axis of symmetry is $\left(0 , - 4\right)$

Then the coordinates of the vertex $\frac{0 + 0}{2} , \frac{4 - 4}{2} = \left(0 , 0\right)$

Vertex $\left(0 , 0\right)$. It is a point on the parabola.

As we know the vertex, we can form the equation.

The equation of the parabola which opens up and whose vertex is at origin is ${x}^{2} = 4 a y$

We know $a$ is the distance between focus and vertex.

$a = 4$

then, the equation is -

${x}^{2} = 16 y$