College Algebra Problems With Answers sample 10 : Equation of Hyperbola
College algebra problems on the equations of hyperbolas are presented.
Detailed solutions are at the bottom of the page.
Problems
Problem 1
Find the transverse axis, the center, the foci and the vertices of the hyperbola whose equation is
\(\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1\)
Problem 2
Find the transverse axis, the center, the foci and the vertices of the hyperbola whose equation is
\(16y^2 - x^2 = 16\)
Problem 3
Find the equation of a hyperbola that has the \(y\) axis as the transverse axis, a center at \((0 , 0)\) and passes through the points \((0 , 5)\) and \((2 , 5\sqrt{2})\).
Problem 4
Find the equation of a hyperbola whose vertices are at \((0 , -3)\) and \((0 , 3)\) and has a focus at \((0 , 5)\).
Problem 5
Find the asymptotes of the parabolas given by the equations:
a) \(\dfrac{x^2}{4} - \dfrac{y^2}{36} = 1\)
b) \(y^2 - 49x^2 = 49\)
Problem 6
Find the equation of a hyperbola with vertices at \((0 , -7)\) and \((0 , 7)\) and asymptotes given by the equations \(y = 3x\) and \(y = - 3x\).
Problem 7
Find the equation of a hyperbola with foci at \((-2 , 0)\) and \((2 , 0)\) and asymptotes given by the equation \(y = x\) and \(y = -x\).
Problem 8
Write the equation of a hyperbola with foci at \((-1 , 0)\) and \((1 , 0)\) and one of its asymptotes passes through the point \((1 , 3)\).
Problem 9
Write the equation of a hyperbola with the \(x\) axis as its transverse axis, point \((3 , 1)\) lies on the graph of this hyperbola and point \((4 , 2)\) lies on the asymptote of this hyperbola.
Problem 10
Find the equation of each parabola shown below. The graphs in b) and c) also shows the asymptotes.
a)
.
Figure 1. Graph of hyperbola a)
b)
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Figure 2. Graph of hyperbola b)
c)
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Figure 3. Graph of hyperbola c)
Solutions to the Above Problems
Solution to Problem 1
Transverse axis: \(x\) axis or \(y = 0\)
Center at \((0 , 0)\)
Vertices at \((2 , 0)\) and \((-2 , 0)\)
\(c^2 = 4 + 9 = 13\). Foci are at \((\sqrt{13} , 0)\) and \((-\sqrt{13} , 0)\).
Solution to Problem 2
Divide all terms of the given equation by 16 which becomes \(y^2 - \dfrac{x^2}{16} = 1\)
Transverse axis: \(y\) axis or \(x = 0\)
Center at \((0 , 0)\)
Vertices at \((0 , 1)\) and \((0 , -1)\)
\(c^2 = 1 + 16 = 17\). Foci are at \((0 , \sqrt{17})\) and \((0 , -\sqrt{17})\).
Solution to Problem 3
Since the \(y\) axis is the transverse axis, the equation has the form \(\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1\
)
Use the point \((0 , 5)\) to write: \(5^2 / a^2 = 1\) and find \(a^2 = 25\). Use the second point to write \((5\sqrt{2})^2 / 25 - 2^2 / b^2 = 1\) and find \(b^2 = 4\)
The equation is given by: \(\dfrac{y^2}{25} - \dfrac{x^2}{4} = 1\)
Solution to Problem 4
Since the vertices are at \((0 ,-3)\) and \((0 , 3)\), the transverse axis is the \(y\) axis and the center is at \((0,0)\). The equation has the form: \(\dfrac{y^2}{9} - \dfrac{x^2}{b^2} = 1\), \(a^2 = 9\).
The focus is at \((0 , 5)\) hence \(c = 5\). We now use the formula \(c^2 = a^2 + b^2\) to find \(b^2 = 25 - 9 = 16\)
The equation may be written as: \(\dfrac{y^2}{9} - \dfrac{x^2}{16} = 1\)
Solution to Problem 5
a) \(y = 3x\) and \(y = -3x\)
b) \(y = 7x\) and \(y = -7x\)
Solution to Problem 6
Since the vertices are at \((0 , -7)\) and \((0 , 7)\), the transverse axis of the hyperbola is the \(y\) axis, the center is at \((0 , 0)\) and the equation of the hyperbola has the form \(\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1\) with \(a^2 = 49\). The asymptote is given by \(y = \pm \dfrac{a}{b}x\), hence \(\dfrac{a}{b} = 3\) which gives \(a^2 = 9 b^2\).
Since \(a^2 = 49\), \(9 b^2 = 49\) and \(b^2 = \dfrac{49}{9}\)
The equation of the hyperbola is given by: \(\dfrac{y^2}{49} - \dfrac{9x^2}{49} = 1\)
Solution to Problem 7
Since the foci are at \((-2 , 0)\) and \((2 , 0)\), the transverse axis of the hyperbola is the \(x\) axis, the center is at \((0 , 0)\) and the equation of the hyperbola has the form \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) with \(c^2 = 4 = a^2 + b^2\)
The asymptote is given by \(y = \pm \dfrac{b}{a}x\), hence \(\dfrac{a}{b} = 1\) which gives \(a^2 = b^2\).
Solve the two equations \(4 = a^2 + b^2\) and \(a^2 = b^2\) to find: \(a^2 = 2\) and \(b^2 = 2\).
The equation of the hyperbola is give by: \(\dfrac{x^2}{2} - \dfrac{y^2}{2} = 1\)
Solution to Problem 8
Since the foci are at \((-1 , 0)\) and \((1 , 0)\), the transverse axis of the hyperbola is the \(x\) axis, the center is at \((0,0)\) and the equation of the hyperbola has the form \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) with \(c^2 = 1^2 = a^2 + b^2\)
The asymptote is given by \(y = \dfrac{b}{a}x\), hence \(\dfrac{a}{b} = \dfrac{3}{1} = 3\) which gives \(a^2 = 9 b^2\).
Solve both equations: \(1 = a^2 + b^2\) and \(a^2 = 9 b^2\).
Solve to find: \(b^2 = \dfrac{1}{10}\) and \(a^2 = \dfrac{9}{10}\)
The equation of the hyperbola is given by: \(\dfrac{10}{9}x^2 - \dfrac{10}{9}y^2 = 1\)
Solution to Problem 9
The equation of the hyperbola has the form: \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\)
Use point \((3 , 1)\) to write: \(\dfrac{3^2}{a^2} - \dfrac{1^2}{b^2} = 1\)
The asymptote has the form: \(y = \pm \dfrac{b}{a}x\), using the point \((4,2)\) that lies on the asymptote we write: \(\dfrac{b}{a} = \dfrac{2}{4} = \dfrac{1}{2}\) or \(4b^2 = a^2\)
Solve the two equations to find: \(a^2 = 5\) and \(b^2 = \dfrac{5}{4}\)
The equation of the hyperbola has the form: \(\dfrac{x^2}{5} - \dfrac{y^2}{\dfrac{5}{4}} = 1\)
Solution to Problem 10
a)
Vertices at \((-1 , 0)\) and \((1 , 0)\) and point \((-3 , 2)\) lies on the hyperbola.
Equation: \(x^2 - \dfrac{y^2}{0.5} = 1\)
b)
Vertices at \((-2 , 0)\) and \((2 , 0)\) and point \((2 , 2)\) lies on one asymptote.
Equation: \(\dfrac{x^2}{4
} - \dfrac{y^2}{4} = 1\)
c)
Vertices at \((0 , 0.5)\) and \((0 , -0.5)\) and asymptote \(y = \dfrac{x}{6}\).
Equation: \(\dfrac{y^2}{0.25} - \dfrac{x^2}{9} = 1\)
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