AP EAMCET · Maths · Hyperbola
The equation of hyperbola whose eccentricity is \(\frac{5}{3}\) and distance between the foci is 10 units is
- A \(16 x^2-9 y^2=16\)
- B \(16 x^2-9 y^2=9\)
- C \(16 x^2-9 y^2=-144\)
- D \(16 x^2-9 y^2=144\)
Answer & Solution
Correct Answer
(D) \(16 x^2-9 y^2=144\)
Step-by-step Solution
Detailed explanation
Given, \(e=5 / 3\) and \(2 a e=10\) \(\Rightarrow \quad 2 a\left(\frac{5}{3}\right)=10 \Rightarrow a=3\) \(\Rightarrow \quad(a e)^2=a^2+b^2 \Rightarrow(5)^2=3^2+b^2\) \(\Rightarrow b^2=25-9=16\) \(\therefore\) Equation of hyperbola is \(\frac{x^2}{9}-\frac{y^2}{16}=1\)…
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