College Algebra Questions and Problems With Solutions
sample 9 : Equation of Parabola

College algebra questions and problems on the equation of parabolas are presented. The solutions with explanations are located at the bottom of the page.

Question 1

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

Question 2

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).

Question 3

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

Question 4

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

Question 5

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

Question 6

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

Question 7

Write an equation for the parabola shown in the graph below and find the focus of the parabola.

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Question 8

Find an equation of the parabola with focus at (8 , 0) and directrix given by the equation x = 2.

Question 9

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.

Question 10

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

Solution Question 1

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

Solution Question 2

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

Solution Question 3

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)

Solution Question 4

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

Solution Question 5

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

Solution Question 6

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)

Solution Question 7

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)

Solution Question 8

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 2|a| = 6, |a| = 3 and since the parabola opens to the right a = 3.
y ² = 12 (x - 5)

Solution Question 9

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 2|a|, 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 2|a|, 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)

Solution Question 10

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).

More References and links

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