Chapter 8
College Algebra · 401 exercises
Problem 57
For the following exercises, determine the value of \(k\) based on the given equation. Given \(2 x^{2}+k x y+12 y^{2}+10 x-16 y+28=0\) find \(k\) for the graph to be an ellipse.
5 step solution
Problem 57
For the following exercises, find the area of the ellipse. The area of an ellipse is given by the formula Area \(=a \cdot b \cdot \pi\). $$ \frac{(x-3)^{2}}{9}+\frac{(y-3)^{2}}{16}=1 $$
4 step solution
Problem 57
Express the equation for the hyperbola as two functions, with y as a function of x. Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. \(\frac{y^{2}}{9}-\frac{x^{2}}{1}=1\)
7 step solution
Problem 58
Recall from Rotation of Axes that equations of conics with an \(x y\) term have rotated graphs. For the following exercises, express each equation in polar form with \(r\) as a function of \(\theta\). $$ 2 x^{2}+4 x y+2 y^{2}=9 $$
5 step solution
Problem 58
Express each equation in polar form with \(r\) as a function of \(\theta\). $$ 2 x^{2}+4 x y+2 y^{2}=9 $$
8 step solution
Problem 58
For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation. $$ V(1,2), \text { Endpoints }(-5,5),(7,5) $$
5 step solution
Problem 58
For the following exercises, find the ellipse. The area of an ellipse is given by the formula Area \(=a \cdot b \cdot \pi\) $$ \frac{(x+6)^{2}}{16}+\frac{(y-6)^{2}}{36}=1 $$
3 step solution
Problem 58
For the following exercises, express the equation for the hyperbola as two functions, with \(y\) as a function of \(x .\) Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. $$ \frac{(x-2)^{2}}{16}-\frac{(y+3)^{2}}{25}=1 $$
5 step solution
Problem 58
For the following exercises, determine the value of \(k\) based on the given equation. Given \(3 x^{2}+k x y+4 y^{2}-6 x+20 y+128=0\) find \(k\) for the graph to be a hyperbola.
5 step solution
Problem 58
For the following exercises, find the area of the ellipse. The area of an ellipse is given by the formula Area \(=a \cdot b \cdot \pi\). $$ \frac{(x+6)^{2}}{16}+\frac{(y-6)^{2}}{36}=1 $$
3 step solution
Problem 58
Express the equation for the hyperbola as two functions, with y as a function of x. Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. \(\frac{(x-2)^{2}}{16}-\frac{(y+3)^{2}}{25}=1\)
5 step solution
Problem 59
Recall from Rotation of Axes that equations of conics with an \(x y\) term have rotated graphs. For the following exercises, express each equation in polar form with \(r\) as a function of \(\theta\). $$ 16 x^{2}+24 x y+9 y^{2}=4 $$
5 step solution
Problem 59
Express each equation in polar form with \(r\) as a function of \(\theta\). $$ 16 x^{2}+24 x y+9 y^{2}=4 $$
7 step solution
Problem 59
For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation. $$ V(-3,-1), \text { Endpoints }(0,5),(0,-7) $$
4 step solution
Problem 59
For the following exercises, express the equation for the hyperbola as two functions, with \(y\) as a function of \(x .\) Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. $$ -4 x^{2}-16 x+y^{2}-2 y-19=0 $$
6 step solution
Problem 59
For the following exercises, find the area of the ellipse. The area of an ellipse is given by the formula Area \(=a \cdot b \cdot \pi\). $$ \frac{(x+1)^{2}}{4}+\frac{(y-2)^{2}}{5}=1 $$
4 step solution
Problem 59
Express the equation for the hyperbola as two functions, with y as a function of x. Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. \(-4 x^{2}-16 x+y^{2}-2 y-19=0\)
5 step solution
Problem 60
Recall from Rotation of Axes that equations of conics with an \(x y\) term have rotated graphs. For the following exercises, express each equation in polar form with \(r\) as a function of \(\theta\). $$ 2 x y+y=1 $$
4 step solution
Problem 60
Express each equation in polar form with \(r\) as a function of \(\theta\). $$ 2 x y+y=1 $$
3 step solution
Problem 60
For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation. $$ V(4,-3), \text { Endpoints }\left(5,-\frac{7}{2}\right),\left(3,-\frac{7}{2}\right) $$
6 step solution
Problem 60
For the following exercises, find the ellipse. The area of an ellipse is given by the formula Area \(=a \cdot b \cdot \pi\) $$ 4 x^{2}-8 x+9 y^{2}-72 y+112=0 $$
6 step solution
Problem 60
For the following exercises, express the equation for the hyperbola as two functions, with \(y\) as a function of \(x .\) Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. $$ 4 x^{2}-24 x-y^{2}-4 y+16=0 $$
7 step solution
Problem 60
For the following exercises, determine the value of \(k\) based on the given equation. Given \(6 x^{2}+12 x y+k y^{2}+16 x+10 y+4=0\) find \(k\) for the graph to be an ellipse.
5 step solution
Problem 60
For the following exercises, find the area of the ellipse. The area of an ellipse is given by the formula Area \(=a \cdot b \cdot \pi\). $$ 4 x^{2}-8 x+9 y^{2}-72 y+112=0 $$
7 step solution
Problem 60
Express the equation for the hyperbola as two functions, with y as a function of x. Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. \(4 x^{2}-24 x-y^{2}-4 y+16=0\)
6 step solution
Problem 61
The mirror in an automobile headlight has a parabolic cross-section with the light bulb at the focus. On a schematic, the equation of the parabola is given as \(x^{2}=4 y .\) At what coordinates should you place the light bulb?
6 step solution
Problem 61
For the following exercises, find the ellipse. The area of an ellipse is given by the formula Area \(=a \cdot b \cdot \pi\) $$ 9 x^{2}-54 x+9 y^{2}-54 y+81=0 $$
5 step solution
Problem 61
For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. The hedge will follow the asymptotes \(y=x\) and \(y=-x,\) and its closest distance to the center fountain is 5 yards.
4 step solution
Problem 62
Find the equation of the ellipse that will just fit inside a box that is 8 units wide and 4 units high.
4 step solution
Problem 62
For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. The hedge will follow the asymptotes \(y=2 x\) and \(y=-2 x,\) and its closest distance to the center fountain is 6 yards.
7 step solution
Problem 62
A hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. The hedge will follow the asymptotes \(y=2 x\) and \(y=-2 x\) , and its closest distance to the center fountain is 6 yards.
5 step solution
Problem 63
A satellite dish is shaped like a paraboloid of revolution. Th s means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed?
3 step solution
Problem 63
Find the equation of the ellipse that will just fit inside a box that is four times as wide as it is high. Express in terms of \(h\) , the height.
6 step solution
Problem 63
For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. The hedge will follow the asymptotes \(y=\frac{1}{2} x\) and \(y=-\frac{1}{2} x,\) and its closest distance to the center fountain is 10 yards.
6 step solution
Problem 64
For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. The hedge will follow the asymptotes \(y=\frac{2}{3} x\) and \(y=-\frac{2}{3} x,\) and its closest distance to the center fountain is 12 yards.
6 step solution
Problem 64
An arch has the shape of a semi-ellipse (the top half of an ellipse). The arch has a height of 8 feet and a span of 20 feet. Find an equation for the ellipse, and use that to find the height to the nearest 0.01 foot of the arch at a distance of 4 feet from the center.
4 step solution
Problem 64
A hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. The hedge will follow the asymptotes \(y=\frac{2}{3} x\) and \(y=-\frac{2}{3} x,\) and its closest distance to the center fountain is 12 yards.
5 step solution
Problem 65
For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. The hedge will follow the asymptotes \(y=\frac{3}{4} x\) and \(y=-\frac{3}{4} x,\) and its closest distance to the center fountain is 20 yards.
5 step solution
Problem 65
An arch has the shape of a semi-ellipse. The arch has a height of 12 feet and a span of 40 feet. Find an equation for the ellipse, and use that to find the distance from the center to a point at which the height is 6 feet. Round to the nearest hundredth.
5 step solution
Problem 65
A searchlight is shaped like a paraboloid of revolution. A light source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 3 feet across, find the depth.
4 step solution
Problem 65
A hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. The hedge will follow the asymptotes \(y=\frac{3}{4} x\) and \(y=-\frac{3}{4} x,\) and its closest distance to the center fountain is 20 yards.
5 step solution
Problem 66
A bridge is to be built in the shape of a semi- elliptical arch and is to have a span of 120 feet. The height of the arch at a distance of 40 feet from the center is to be 8 feet. Find the height of the arch at its center.
6 step solution
Problem 66
A bridge is to be built in the shape of a semielliptical arch and is to have a span of 120 feet. The height of the arch at a distance of 40 feet from the center is to be 8 feet. Find the height of the arch at its center.
6 step solution
Problem 67
An arch is in the shape of a parabola. It has a span of 100 feet and a maximum height of 20 feet. Find the equation of the parabola, and determine the height of the arch 40 feet from the center.
6 step solution
Problem 67
A person in a whispering gallery standing at one focus of the ellipse can whisper and be heard by a person standing at the other focus because all the sound waves that reach the ceiling are reflected to the other person. If a whispering gallery has a length of 120 feet, and the foci are located 30 feet from the center, find the height of the ceiling at the center.
4 step solution
Problem 67
assume an object enters our solar system and we want to graph its path on a coordinate system with the sun at the origin and the x-axis as the axis of symmetry for the object's path. Give the equation of the flight path of each object using the given information. The object enters along a path approximated by the line \(y=2 x-2\) and passes within 0.5 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line \(y=-2 x+2\)
6 step solution
Problem 69
An object is projected so as to follow a parabolic path given by \(y=-x^{2}+96 x,\) where \(x\) is the horizontal distance traveled in feet and \(y\) is the height. Determine the maximum height the object reaches.
4 step solution
Problem 69
For the following exercises, assume an object enters our solar system and we want to graph its path on a coordinate system with the sun at the origin and the \(x\) -axis as the axis of symmetry for the object's path. Give the equation of the flight path of each object using the given information. The object enters along a path approximated by the line \(y=\frac{1}{3} x-1\) and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line \(y=-\frac{1}{3} x+1\).
7 step solution
Problem 69
assume an object enters our solar system and we want to graph its path on a coordinate system with the sun at the origin and the x-axis as the axis of symmetry for the object's path. Give the equation of the flight path of each object using the given information. The object enters along a path approximated by the line \(y=\frac{1}{3} x-1\) and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line \(y=-\frac{1}{3} x+1 .\)
8 step solution
Problem 70
For the object from the previous exercise, assume the path followed is given by \(y=-0.5 x^{2}+80 x\). Determine how far along the horizontal the object traveled to reach maximum height.
4 step solution