sample 10 : Equation of Hyperbola

College algebra problems on the equations of hyperbolas are presented. Detailed solutions are at the bottom of the page.

\(\dfrac{x^2}{4} - \dfrac{y^2}{9} = 1\)

\(16y^2 - x^2 = 16\)

a) \(\dfrac{x^2}{4} - \dfrac{y^2}{36} = 1\)

b) \(y^2 - 49x^2 = 49\)

a) b) c)

Center at \((0 , 0)\)

Vertices at \((2 , 0)\) and \((-2 , 0)\)

\(c^2 = 4 + 9 = 13\). Foci are at \((\sqrt{13} , 0)\) and \((-\sqrt{13} , 0)\).

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 \((0 , -\sqrt{17})\).

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: \(\dfrac{y^2}{25} - \dfrac{x^2}{4} = 1\)

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: \(\dfrac{y^2}{9} - \dfrac{x^2}{16} = 1\)

b) \(y = 7x\) and \(y = -7x\)

Since \(a^2 = 49\), \(9 b^2 = 49\) and \(b^2 = \dfrac{49}{9}\)

The equation of the hyperbola is given by: \(\dfrac{y^2}{49} - \dfrac{9x^2}{49} = 1\)

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: \(\dfrac{x^2}{2} - \dfrac{y^2}{2} = 1\)

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: \(\dfrac{10}{9}x^2 - \dfrac{10}{9}y^2 = 1\)

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: \(\dfrac{x^2}{5} - \dfrac{y^2}{\dfrac{5}{4}} = 1\)

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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