# 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.

## Questions

### 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
$$x^2 = 4ay = 16y$$

### 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
$$y^2 = 4ax$$
The point $$(-2, 4)$$ lies on the parabola: $$4^2 = 4a(-2)$$ Solve for $$a$$: $$a = -2$$ ; the equation is: $$y^2 = -8x$$

### 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 $$x = h - a$$.
$$2(y^2 + 4y) + x + 1 = 0$$
$$2((y + 2)^2 - 4) + x + 1 = 0$$
$$(y + 2)^2 = -\frac{1}{2}(x - 7)$$
Vertex at $$(7, - 2)$$
$$-\frac{1}{2} = 4a$$, hence $$a = -\frac{1}{8}$$
Focus at $$(7 - \frac{1}{8}, -2) = (6.875, -2)$$
Axis of symmetry given by $$y = -2$$
Directrix is a vertical line given by: $$x = h - a = 7 + \frac{1}{8} = 7.125$$

### 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 $$a = \frac{1}{4}$$
Focus at $$(4, -14 + \frac{1}{4}) = (4, -13.75)$$
Axis of symmetry given by $$x = 4$$
Directrix is a horizontal line given by: $$y = k - a = -14 - \frac{1}{4} = -14.25$$

### 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 given 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

For the equation of the parabola of the form $$(y - k)^2 = 4a(x - h)$$ the vertex is at $$(h, k)$$
From the given graph, the vertex is at $$(h, k) = (2, 4)$$, therefore the equation of the parabola is given 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 line $$x = 2$$ 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 is 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 $$a = 3$$. The equation of the parabola is given by: $$y^2 = 12(x - 5)$$

### 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 particularly point $$(-6, -8)$$, should be at equal distances from the directrix and the focus. Hence
$$10 = \sqrt{(6^2 + (b + 8))^2}$$
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 the 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 the 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: $$x^2 = 4ay$$
Point $$(100, 50)$$ lies on the graph of the parabolic dish, hence $$100^2 = 4a \times 50$$
Solve to find: $$a = 50$$ which is also the distance from the vertex at $$(0, 0)$$ to the focus. Hence the focus is at $$(0, 50 \, \text{cm})$$.