# Question #75d40

##### 1 Answer

#### Answer:

Focus of the parabola is

#x^2=16y#

#### Explanation:

Given -

#12(x-3)=(y+1)^2#

It can be written as -

#(y+1)^2=12(x-3)#

Since it is in the form

This is in the form-

#(y-k)^2=4a(x-h)#

Where -

#(h,k)# is vertex

#a# is distance of focus or directrix from the vertex.

From the given equation we can find the value of

#(y+1)^2=4xx3(x-3)^3#

Then

Vertex

focus

PROBLEM 2

Given -

Focus

Directrix

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

Then the coordinates of the vertex

Vertex

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

We know

#a=4#

then, the equation is -

#x^2=16y#