AP EAMCET · Maths · Hyperbola
If \(e_1\) and \(e_2\) are respectively the eccentricities of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) and its conjugate hyperbola, then the line \(\frac{x}{2 e_1}+\frac{y}{2 e_2}=1\) touches the circle having centre at the origin, then its radius is
- A 2
- B \(e_1+e_2\)
- C \(e_1 e_2\)
- D 4
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
Hyperbola : \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 ; e_1=\sqrt{\frac{a^2+b^2}{a^2}}\) Conjugate Hyperbola : \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1 ; e_2=\sqrt{\frac{a^2+b^2}{b^2}}\) Line:…
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