AP EAMCET · Maths · Hyperbola
The area of the quadrilateral formed with the foci of the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) and its conjugate hyperbola is (in square units)
- A 24
- B 16
- C 25
- D 50
Answer & Solution
Correct Answer
(D) 50
Step-by-step Solution
Detailed explanation
Hyperbola : \(\frac{x^2}{16}-\frac{y^2}{9}=1 \Rightarrow e_1=\sqrt{1+\frac{9}{16}}=\frac{5}{4}\) Conjugate hyperbola : \(\frac{x^2}{9}-\frac{y^2}{16}=1 \Rightarrow e_2=\sqrt{1+\frac{16}{9}}=\frac{5}{3}\) \(\Rightarrow e_2 \gt e_1\) \(\Rightarrow\) Area of quadrilateral…
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