The eccentricity of the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) is
A. \(\frac53\)
B. \(\frac{\sqrt7}{4}\)
C. \(\frac45\)
D. \(\frac54\)
1 answer
Best answer
Answer: D. \(\frac54\)
\(e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac9{16}}=\frac54\).
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· 23 Oct 2026
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