Question #75d40

1 Answer
Jan 4, 2018

Answer:

Focus of the parabola is #(6, -1)#

#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 #y^2=4ax#; it opens to the right

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 #a#. For this, we have to rewrite the equation like this.

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

Then #a=3#
Vertex #(3,-1)#

focus #(6, -1)#

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PROBLEM 2

Given -

Focus #(0, 4)#
Directrix #y=-4#

Look at the graph

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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 #(0, -4)#

Then the coordinates of the vertex #(0+0)/2, (4-4)/2= (0, 0)#

Vertex #(0, 0)#. 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=4ay#

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

#a=4#

then, the equation is -

#x^2=16y#