
Find an equation of the parabola with focus at (0 , 4) and vertex at (0 , 0).

Find an equation of the parabola with vertex at (0 , 0), the x axis is its axis of symmetry and its graph contains the point (2 , 4).

Find an equation of the parabola with vertex at (0 , 2) and focus at (0 , 6).

Find the vertex, the focus, the axis of symmetry and the directrix of the parabola defined by the equation
2y^{ 2} + 8y + x + 1 = 0

Find the vertex, the focus, the axis of symmetry and the directrix of the parabola defined by the equation
x^{ 2}  8x  y + 2 = 0

Find an equation of the parabola with vertex at (2 , 2) and focus at (2 , 8).

Write an equation for the parabola shown in the graph below and find the focus of the parabola.
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Find an equation of the parabola with focus at (8 , 0) and directrix given by the equation x = 2.

Find an equation of the parabola with directrix given by the equation y = 2, a focus on the y axis, and the point (6 , 8) lies on the parabola.

A parabolic dish with a diameter of 200 cm and a maximum depth of 50 cm is shown below. Find the focus of the dish.
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Solutions to the Above Questions

The distance from the vertex (0 , 0) to the focus (0 , 4) is a = 4. Since the vertex is at (0 , 0) and the focus is at (0 , 4) on the y axis, the parabola opens upward which means a = 4 and its equation is given by
x^{ 2} = 4a y = 16 y

Since the x axis is the axis of symmetry of the parabola and its vertex is at the origin, the equation of the parabola has the form
y^{ 2} = 4a x
The point (2 , 4) lies on the parabola: 4^{ 2} = 4a (2)
Solve for a: a = 2 ; the equation is: y^{ 2} = 8 x

Find an equation of the parabola with vertex at (0 , 2) and focus at (0 , 6).
Since the vertex is at (0 , 2) and the focus is at (0 , 6), the parabola opens upward and the equation of a parabola with vertex at (h , k) is given by
(x  h)^{ 2} = 4a (y  k)
h = 0 and k = 2; the equation is given by: x^{ 2} = 4a (y  2)
a = distance from vertex to focus = 4 , since parabola opens upward a = 4
the equation is given by: x^{ 2} = 16 (y  2)

We first complete the square using the terms in y and y^{ 2} and write the given equation in the form (y  k)^{ 2} = 4a (x  h) where (h , k) is the vertex and the focus is at (h + a , k), the axis of symmetry is given by y = k and the directrix is given by x = h  a.
2(y^{ 2} + 4y) + x + 1 = 0
2((y + 2)^{ 2}  4) + x + 1 = 0
(y + 2)^{ 2}=  (1/2)(x  7)
vertex at (7 , 2)
(1/2) = 4a , hence a = 1/8
focus at (7  1/8 , 2) = (6.875 , 2)
axis of symmetry given by y = 2
Directrix is a vertical line given by: x = h  a = 7 + 1/8 = 7.125

We first complete the square using the terms in x and x^{ 2} and write the given equation in the form (x  h)^{ 2} = 4a (y  k) where (h , k) is the vertex and the focus is at (h , k + a), the axis of symmetry is given by x = h and the directrix is given by y = k  a
x^{ 2}  8x  y + 2 = 0
((x  4)^{ 2}  16)  y + 2 = 0
(x  4)^{ 2}= (y + 14)
vertex at (4 , 14)
1 = 4a , hence a = 1/4
focus at (4 , 14 + 1/4) = (4 , 13.75)
axis of symmetry given by x = 4
Directrix is a horizontal line given by: y = k  a =  14  1/4 = 14.25

The vertex and the focus are on the same vertical line x = 2 with the focus below the vertex, therefore the parabola opens downward and its equation has the form
(x  h)^{ 2} = 4a (y  k) , vertex at (h , k)
with the vertex at (h , k) = (2, 2), the equation is give by
(x + 2)^{ 2} = 4a (y + 2)
The distance from the vertex (2 , 2) to the focus (2 , 8) is a = 6. Since the parabola opens downward a = 6.
The equation of the parabola is given by: (x + 2)^{ 2} = 24 (y + 2)

The vertex is at (2 , 4)
(y  k)^{ 2} = 4a (x  h) , vertex at (h , k)
with the vertex at (h , k) = (2, 4), the equation is give by
(y  4)^{ 2} = 4a (x  2)
Use the point (2 , 2) to find a as follows
(2  4)^{ 2} = 4a (2  2)
solve the above for a: a = 2.25
Equation of the parabola: (y  4)^{ 2} = 9 (x  2)
The focus is at (h + a , k) = (2  2.25 , 4) = (0.25 , 4)

Since the directrix is the x = 2 and the focus is at (8 , 0), the parabola has the x axis as an axis of symmetry and opens to the right. Its equation is of the form
(y  k)^{ 2} = 4a (x  h) , vertex at (h , k)
The vertex is the midpoint of the point (2 , 0), which the point of intersection of the directrix and the x axis, and the focus ( 8 , 0). Hence h = 5 and k = 0. The distance between the directrix and the focus is twice a. Hence 2a = 6, a = 3 and since the parabola opens to the right a = 3.
y^{ 2} = 12 (x  5)

The focus is on the y axis and therefore has the form F(0 , b). The distance from the given point to the directrix is equal to 10. By definition any point on the parabola, and particular point ( 6 , 8), should be at equal distances from the directrix and the focus. Hence
10 = √(6^{ 2} + (b + 8))^{ 2})
Solve the above for b: 2 solutions b = 0 and b = 16
THERE ARE TWO SOLUTIONS TO THIS PROBLEM
First solution: the distance from the focus F(0 , 0) to the directrix y = 2 is equal to 2a, hence a = 1 and a = 1 since parabola opens downward. The vertex is at (0 , 1), at equal distances from the focus and the directrix.
equation: x^{ 2} =  4 (y  1)
Second solution: the distance from the focus F(0 , 16) to the directrix y = 2 is equal to 2a, hence a = 9 and a = 9 since parabola opens downward. The vertex is at (0 , 7), at equal distances from the focus and the directrix.
equation: x^{ 2} =  36 (y + 7)

The equation of the parabolic dish is of the form: x^{ 2} = 4 a y
Point (100 , 50) lies on the graph of the parabolic dish, hence
100^{ 2} = 4 a * 50
solve to find: a = 50 which is also the distance from the vertex at (0 , 0) to the focus. Hence the focus is at (0 , 50 cm).
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