What is the graph and equation of a parabola with

Question

the focus of (5,0) and vertex of (5,2)?

Answer 1

Given that vertex (5 , 2) and focus (5 , 0)

Equation of a parabola if focus and vertex is given is (x - h)^2 = 4p(y - k)

(h , k) are the vertecies and p = distance from vertex to focus

Taken as positive if focus is above the vertex and negitive if focus is below the vertex

p = 0 - 2

p = -2

It is taken as positive because focus is above vertex

Therefore p = 2

4p = 4(2)

4p = 8

Now take general equation of a parabola

(x - h)^2 = 4p(y - k)

Substitute the values (h , k) and 4p

(x - 5)^2 = 8(y - 2)

Therefor the required equation of a parabola is (x - 5)^2 = 8(y - 2)

 

Answer 2

  

The vertex (5,2) and focus (5,0)

Since the x coordinate of the vertex and focus are the same,they are one top of other,

So this is regular vertical parabola where x  part is squred.

p  is distance between vertex and focus and also equals to distance between vertex and directrix.

p  = 0 - 2 = - 2

p  is negitive then this is regular parabola that opens downward.

This means the directrix being on the outside of the parabola is 2 units above the vertex.

The parabola equation is  (x - h )² = 4p (y - k )

Substitute the values (h, k) = (5,2) and 4p = -2 in above equation.

The parabola equation is (x - 5)²= -8(y - 2)

(x - 5)²/-8 = y -2

y  = -1/8(x -5)² + 2

Directrix is y = k - p = 2 - (-2) = 4

Directrix y  = 4

Choose values for x and find the corresponding values for y.

x	y  = -1/8(x -5)^2 +2	(x, y )
1	y  = -1/8(1-5)^2 + 2 y = 0	(1,0)
-2	y  = -1/8(-2-5)^2 +2 y = -33/8	(-2,-33/8)
10	y  = -1/8(10-5)^2+2 y = -9/8	(10,-9/8)
12	y  =-1/8(12-5)^2 +2 y = -33/8	(12,-33/8)

1.Draw a coordinate plane.

2.Plot the directrix,focus and vertex and coordinate points found in the table.

3.Then sketch the graph, connecting the points with a smooth curve.