College Algebra Problems With Answers
Equation of Hyperbola

• Transverse axis: \(x\) axis or \(y = 0\)
• Center: at \((0 , 0)\)
• Vertices: at \((2 , 0)\) and \((-2 , 0)\)
• Foci: \(c^2 = 4 + 9 = 13\), so foci are at \((\sqrt{13} , 0)\) and \((-\sqrt{13} , 0)\).

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 \((0 , -1)\)
• Foci: \(c^2 = 1 + 16 = 17\), so foci are at \((0 , \sqrt{17})\) and \((0 , -\sqrt{17})\).

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: \(\dfrac{5^2}{a^2} = 1\), which gives \(a^2 = 25\).

Use the second point \((2, 5\sqrt{2})\) to write:

\[ \dfrac{(5\sqrt{2})^2}{25} - \dfrac{2^2}{b^2} = 1 \implies \dfrac{50}{25} - \dfrac{4}{b^2} = 1 \implies 2 - 1 = \dfrac{4}{b^2} \implies b^2 = 4 \]

The equation is given by:

\[ \dfrac{y^2}{25} - \dfrac{x^2}{4} = 1 \]

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:

\[ \dfrac{y^2}{9} - \dfrac{x^2}{b^2} = 1 \quad (\text{with } a^2 = 9) \]

The focus is at \((0 , 5)\), hence \(c = 5\). We use the formula \(c^2 = a^2 + b^2\) to find \(b^2\):

\[ 5^2 = 9 + b^2 \implies 25 = 9 + b^2 \implies b^2 = 16 \]

The equation may be written as:

\[ \dfrac{y^2}{9} - \dfrac{x^2}{16} = 1 \]

• a) \(y = 3x\) and \(y = -3x\)
• b) Dividing by 49 gives \(y^2 - \dfrac{x^2}{1^2} = 1\), yielding asymptotes \(y = 7x\) and \(y = -7x\).

Since the vertices are at \((0 , -7)\) and \((0 , 7)\), the transverse axis is the \(y\) axis, the center is at \((0 , 0)\), and the equation has the form:

\[ \dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1 \quad \text{with } a^2 = 49 \]

The asymptotes are given by \(y = \pm \dfrac{a}{b}x\), hence \(\dfrac{a}{b} = 3\), which gives \(a = 3b \implies a^2 = 9b^2\).

Since \(a^2 = 49\), we have \(9b^2 = 49 \implies b^2 = \dfrac{49}{9}\).

The equation of the hyperbola is given by:

\[ \dfrac{y^2}{49} - \dfrac{9x^2}{49} = 1 \]

Since the foci are at \((-2 , 0)\) and \((2 , 0)\), the transverse axis is the \(x\) axis, the center is at \((0 , 0)\), and the equation has the form:

\[ \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \quad \text{with } c^2 = 2^2 = 4 = a^2 + b^2 \]

The asymptotes are given by \(y = \pm \dfrac{b}{a}x\), hence \(\dfrac{b}{a} = 1\), which gives \(a^2 = b^2\).

Substitute \(a^2 = b^2\) into \(4 = a^2 + b^2\):

\[ 4 = a^2 + a^2 \implies 2a^2 = 4 \implies a^2 = 2 \text{ and } b^2 = 2 \]

The equation of the hyperbola is given by:

\[ \dfrac{x^2}{2} - \dfrac{y^2}{2} = 1 \]

Since the foci are at \((-1 , 0)\) and \((1 , 0)\), the transverse axis is the \(x\) axis, the center is at \((0,0)\), and the equation has the form:

\[ \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \quad \text{with } c^2 = 1^2 = 1 = a^2 + b^2 \]

The asymptote passing through the origin and \((1, 3)\) has slope \(3\), so \(\dfrac{b}{a} = 3 \implies b = 3a \implies b^2 = 9a^2\).

Substitute \(b^2 = 9a^2\) into \(1 = a^2 + b^2\):

\[ 1 = a^2 + 9a^2 \implies 10a^2 = 1 \implies a^2 = \dfrac{1}{10} \text{ and } b^2 = \dfrac{9}{10} \]

The equation of the hyperbola is given by:

\[ 10x^2 - \dfrac{10}{9}y^2 = 1 \]

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:

\[ \dfrac{3^2}{a^2} - \dfrac{1^2}{b^2} = 1 \implies \dfrac{9}{a^2} - \dfrac{1}{b^2} = 1 \]

The asymptote has the form \(y = \pm \dfrac{b}{a}x\). Using the point \((4, 2)\) that lies on the asymptote:

\[ 2 = \dfrac{b}{a}(4) \implies \dfrac{b}{a} = \dfrac{2}{4} = \dfrac{1}{2} \implies a = 2b \implies a^2 = 4b^2 \]

Substitute \(a^2 = 4b^2\) into the first equation:

\[ \dfrac{9}{4b^2} - \dfrac{1}{b^2} = 1 \implies \dfrac{9 - 4}{4b^2} = 1 \implies \dfrac{5}{4b^2} = 1 \implies 4b^2 = 5 \implies b^2 = \dfrac{5}{4} \]

Then \(a^2 = 4\left(\dfrac{5}{4}\right) = 5\).

The equation of the hyperbola is:

\[ \dfrac{x^2}{5} - \dfrac{y^2}{\dfrac{5}{4}} = 1 \]

• a) Equation: \(x^2 - \dfrac{y^2}{0.5} = 1\)
• b) Equation: \(\dfrac{x^2}{4} - \dfrac{y^2}{4} = 1\)
• c) Equation: \(\dfrac{y^2}{0.25} - \dfrac{x^2}{9} = 1\)