College Algebra Problems With Answers sample 8 : Equation of Ellipse
College algebra problems on the equation of the ellipse are presented.
More Problems on ellipses with detailed solutions are included in this site. The solutions to the questions below are at the bottom of the page.
Problems
Problem 1
What is the major axis and its length for the following ellipse?
\[ \dfrac{1}{9}x^{2} + \dfrac{9}{25}y^{2} = \dfrac{1}{25} \]
Problem 2
An ellipse is given by the equation
\[ 8x^{2} + 2y^{2} = 32 \]
Find
a) the major axis and the minor axis of the ellipse and their lengths,
b) the vertices of the ellipse,
c) and the foci of this ellipse.
Problem 3
Find the equation of the ellipse whose center is the origin of the axes and has a focus at \( (0 , -4) \) and a vertex at \( (0 , -6) \).
Problem 4
Find the equation of the ellipse whose foci are at \( (0 , -5) \) and \( (0 , 5) \) and the length of its major axis is 14.
Problem 5
An ellipse has the x axis as the major axis with a length of 10 and the origin as the center. Find the equation of this ellipse if the point \( (3 , \dfrac{16}{5}) \) lies on its graph.
Problem 6
An ellipse has the following equation
\[ 0.2x^{2} + 0.6y^{2} = 0.2 \]
a) Find the equation of part of the graph of the given ellipse that is to the left of the y axis.
b) Find the equation of part of the graph of the given ellipse that is below the x axis.
Problem 7
An ellipse is given by the equation
\[ \dfrac{(x - 1)^{2}}{9} + \dfrac{(y + 4)^{2}}{16} = 1 \]
Find
a) its center,
a) its major and minor axes and their lengths,
b) its vertices,
c) and the foci.
Problem 8
Find the equation of the ellipse whose foci are at \( (-1 , 0) \) and \( (3 , 0) \) and the length of its minor axis is 2.
Problem 9
An ellipse is defined by its parametric equations as follows
\[ x = 6 \sin(t) \quad \text{and} \quad y = 4 \cos(t) \]
Find the center, the major and minor axes and their lengths of this ellipse.
Problem 10
An ellipse is given by the equation
\[ 4x^{2} + 3y^{2} -16x + 18y = -31 \]
Find
a) the center of the ellipse,
b) its major and minor axes and their lengths,
c) its vertices,
d) and its foci.
Solutions to the Above Problems
Solution to Problem 1
Multiply all terms of the equation by 25 to obtain
\[ \dfrac{25}{9}x^{2} + 9y^{2} = 1 \]
The above equation may be written in the form \( \dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1 \) as follows
\[ \dfrac{x^{2}}{\left(\dfrac{3}{5}\right)^{2}} + \dfrac{y^{2}}{\left(\dfrac{1}{3}\right)^{2}} = 1 \]
with \( a = \dfrac{3}{5} \) and \( b = \dfrac{1}{3} \). The major axis is the x axis and its length is equal to \( 2a = \dfrac{6}{5} = 1.2 \)
Solution to Problem 2
a) Divide all terms of the equation by 32 to obtain
\[ \dfrac{x^{2}}{4} + \dfrac{y^{2}}{16} = 1 \]
The above equation may be written as follows
\[ \dfrac{x^{2}}{b^{2}} + \dfrac{y^{2}}{a^{2}} = 1 \]
with \( a = 4 \) and \( b = 2 \) and \( a > b \). Hence the major axis is the y axis and the minor axis is the x axis. The length of the major axis = \( 2a = 8 \) and the length of the minor axis = \( 2b = 4 \)
b) The vertices are on the major axis at the points \( (0 , a) = (0 , 4) \) and \( (0 , -a) = (0 , -4) \)
c) The foci are on the major axis at the points \( (0 , c) \) and \( (0 , -c) \) such that
\( c^{2} = a^{2} - b^{2} = 12 \).
Hence the foci are at the points \( (0 , 2\sqrt{3}) \) and \( (0 , -2\sqrt{3}) \)
Solution to Problem 3
Both the focus and the vertex lie in the y axis which means that the major axis is the y axis. The equation of the ellipse has the form
\[ \dfrac{x^{2}}{b^{2}} + \dfrac{y^{2}}{a^{2}} = 1 \]
\( a \) is the distance from the center of the ellipse to the a vertex and is equal to 6. \( c \) is the distance from the center of the ellipse to a focus and is equal to 4. Also \( a \), \( b \) and \( c \) are related as follows
\( b^{2} = a^{2} - c^{2} = 36 - 16 = 20 \)
\( b = 2\sqrt{5} \)
The equation of the ellipse is given by
\[ \dfrac{x^{2}}{20} + \dfrac{y^{2}}{36} = 1 \]
Solution to Problem 4
From the coordinates of the foci, \( c = 5 \) and the major axis is the y axis. From the length of the major axis, we obtain \( a = 7 \). Also \( b^{2} = a^{2} - c^{2} = 24 \).
The equation of the ellipse is given by
\[ \dfrac{x^{2}}{24} + \dfrac{y^{2}}{49} = 1 \]
Solution to Problem 5
Length of major axis is 10 hence \( a = 5 \) and the equation may be written as follows
\[ \dfrac{x^{2}}{25} + \dfrac{y^{2}}{b^{2}} = 1 \]
We now use the fact that the point \( (3 , \dfrac{16}{5}) \) lies on the graph of the ellipse to find \( b^{2} \).
\[ \dfrac{3^{2}}{25} + \left(\dfrac{16}{5}\right)^{2} \dfrac{1}{b^{2}} = 1 \]
Solve the above for \( b \) to find \( b = 4 \) and write the equation as follows
\[ \dfrac{x^{2}}{25} + \dfrac{y^{2}}{16} = 1 \]
Solution to Problem 6
An ellipse has the following equation
\[ 0.2x^{2} + 0.6y^{2} = 0.2 \]
a) Solve the above equation for \( x \) and select the solution for which \( x \) is positive
\( x = \sqrt{1 - 3y^{2}} \)
b) Solve the ellipse equation for \( y \) and select the solution for which \( y \) is negative.
\( y = - \sqrt{\dfrac{1}{3} - \dfrac{1}{3}x^{2}} \)
Solution to Problem 7
Given the equation
\[ \dfrac{(x - 1)^{2}}{9} + \dfrac{(y + 4)^{2}}{16} = 1 \]
a) Ellipse with center at \( (h , k) = (1 , -4) \) with \( a = 4 \) and \( b = 3 \).
a) its major axis is the line \( x = 1 \), and its minor is the line \( y = -4 \).
length of major axis = \( 2a = 8 \) , length of minor axis = \( 2b = 6 \)
b) vertices at: \( (1 , -4 + 4) = (1 , 0) \) and \( (1 , -4 - 4) = (1 , -8) \)
c) \( c = \sqrt{a^{2} - b^{2}} = \sqrt{7} \)
Foci at: \( (1 , -4 + \sqrt{7}) \) and \( (1 , -4 - \sqrt{7}) \)
Solution to Problem 8
Find the equation of the ellipse whose foci are at \( (-1 , 0) \) and \( (3 , 0) \) and the length of its minor axis is 2.
The center of the ellipse is
the midpoint of the two foci and is at \( (2 , 0) \).
\( c \) is the length from one foci to the center, hence \( c = 2 \).
length of minor axis 2 = \( 2b \) hence \( b = 1 \)
\( a^{2} = b^{2} + c^{2} = 5 \)
Since the foci are on the x axis, the major axis of the ellipse is the x axis.
Equation of ellipse:
\( (x - 2)^{2} / 5 + y^{2} = 1 \)
.
Solution to Problem 9
\[ x = 6 \sin(t) \quad \text{and} \quad y = 4 \cos(t) \]
The parametric equations can be written as follows:
\( x / 6 = \sin(t) \) and \( y / 4 = \cos(t) \)
Square both sides of the two equations: \( (x / 6)^{2} = \sin^{2}(t) \) and \( (y / 4)^{2} = \cos^{2}(t) \)
Use the fact that \( \sin^{2}(t) + \cos^{2}(t) = 1 \) to write
\( x^{2} / 36 + y^{2} / 16 = 1 \)
\( a = 6 \), \( b = 4 \), \( c = \sqrt{36 - 16} = 2\sqrt{5} \)
major axis: x axis , length = 12
minor axis: y axis , length = 8
