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 \( (0, 50 \, \text{cm}) \).
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