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Answer 1 · 2018-01-04T14:02:23

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

$x^2=16y$

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 -

$x^2=16y$