What is the asymptotes equation for the hyperbola x^2 / 16 - y^2 / 9 = 1?
Answer and explanation
Correct answer: A. y = +/-(3/4)x
- Step 1: For a horizontal hyperbola x^2/a^2 - y^2/b^2 = 1, the equations of the asymptotes are y = +/-(b/a)x.
- Step 2: Here a^2 = 16 (a = 4) and b^2 = 9 (b = 3). Thus the asymptotes are y = +/-(3/4)x.
Answer
y = +/-(3/4)x.
Why not the others: 'y = +/-(4/3)x' computes a/b instead of b/a (the vertical hyperbola asymptotes); 'y = +/-(9/16)x' squares the slope coefficients; 'y = +/-(16/9)x' squares the inverted ratio.
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