Question

The distance between the foci of a hyperbola is 16 and its eccentricity is \( \sqrt{2} \). Then its equation is

• A. \( x^2 - y^2 = 32 \)
• B. \( 3x^2 - 2y^2 = 7 \)
• C. \( 2x^2 - 3y^2 = 7 \)
• D. \( \frac{x^2}{4} - \frac{y^2}{9} = 1 \)

Hint:

For hyperbolas, use the relationships \( c^2 = a^2 + b^2 \) and \( e = \frac{c}{a} \) to find the necessary values and form the equation.

Answer 1

The Correct Option is A

The general equation of a hyperbola with horizontal transverse axis is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] where \(2c\) is the distance between the foci, and \(e\) is the eccentricity. We know the relationship: \[ e = \frac{c}{a} \] Also, for a hyperbola, \( c^2 = a^2 + b^2 \). Step 1: Use given data From the problem, we are given that the distance between the foci is 16, so \(2c = 16\), implying that: \[ c = 8 \] Also, the eccentricity \( e = \sqrt{2} \), so: \[ e = \frac{c}{a} \implies \sqrt{2} = \frac{8}{a} \implies a = \frac{8}{\sqrt{2}} = 4\sqrt{2} \] Step 2: Calculate \( b^2 \) Using the relationship \( c^2 = a^2 + b^2 \), we can solve for \( b^2 \): \[ c^2 = a^2 + b^2 \implies 8^2 = (4\sqrt{2})^2 + b^2 \implies 64 = 32 + b^2 \] \[ b^2 = 64 - 32 = 32 \] Step 3: Write the equation Substitute the values of \( a^2 \) and \( b^2 \) into the standard equation of the hyperbola: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \implies \frac{x^2}{32} - \frac{y^2}{32} = 1 \] Multiplying both sides by 32 gives: \[ x^2 - y^2 = 32 \] Thus, the equation of the hyperbola is \( x^2 - y^2 = 32 \).