What are the equations of the asymptotes of the hyperbola x^2 / 16 - y^2 / 9 = 1?
Answer and explanation
Correct answer: D. y = +/- (3/4) x
- Step 1: Standard hyperbola x^2/a^2 - y^2/b^2 = 1 has asymptotes y = +/- (b/a) x. Step 2. Here a^2 = 16 => a = 4, and b^2 = 9 => b = 3. Step 3. Asymptotes: y = +/- (3/4) x. Answer: y = +/- (3/4) x. Why not the others: y = +/- (4/3) x inverts b/a.
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