Question

For a hyperbola, the vertices are at \( (6, 0) \) and \( (-6, 0) \). If the foci are at \( (2\sqrt{10}, 0) \) and \( -2\sqrt{10}, 0) \), then the equation of the hyperbola is:

• A. \(\frac{x^2}{36} - \frac{y^2}{76} = 1\)
• B. \(\frac{x^2}{76} - \frac{y^2}{36} = 1\)
• C. \(\frac{x^2}{6} - \frac{y^2}{2} = 1\)
• D. \(\frac{x^2}{4} - \frac{y^2}{36} = 1\)
• E. \(\frac{x^2}{36} - \frac{y^2}{4} = 1\)

Hint:

For hyperbolas, always ensure to correctly identify whether \(a^2\) or \(b^2\) is associated with the \(x^2\) or \(y^2\) term based on the orientation and length of the axes, and check the relationship \(c^2 = a^2 + b^2\) for any errors.

Answer 1

Answer: (E)

Given that the vertices are at \( (6, 0) \) and \( (-6, 0) \), the length of the transverse axis \(2a\) is \(12\), so \(a = 6\). Therefore, \(a^2 = 36\).
The foci are at \( (2\sqrt{10}, 0) \) and \( (-2\sqrt{10}, 0) \), indicating the distance from the center to each focus \(c = 2\sqrt{10}\). Thus, \(c^2 = 40\).
Using the relationship for a hyperbola, \(c^2 = a^2 + b^2\), we can find \(b^2\): \[ 40 = 36 + b^2 \] \[ b^2 = 4 \] The standard form of the equation of a hyperbola centered at the origin with the transverse axis along the x-axis is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] Substituting \(a^2\) and \(b^2\): \[ \frac{x^2}{36} - \frac{y^2}{4} = 1 \]