TS EAMCET · Maths · Hyperbola
The equation of the hyperbola, whose eccentricity is \(\sqrt{2}\) and whose foci are 16 units apart, is
- A \(9 x^2-4 y^2=36\)
- B \(2 x^2-3 y^2=7\)
- C \(x^2-y^2=16\)
- D \(x^2-y^2=32\)
Answer & Solution
Correct Answer
(D) \(x^2-y^2=32\)
Step-by-step Solution
Detailed explanation
Let the equation of hyperbola be \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) We have, \(e=\sqrt{2} \text { and } 2 a e=16 \Rightarrow 2 a(\sqrt{2})=16\) \(\Rightarrow \quad a=\frac{16}{2 \sqrt{2}}=4 \sqrt{2}\) \(\therefore\) Equation of hyperbola is…
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