AP EAMCET · Maths · Ellipse
The radius of the circle passing through the foci of the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\) and having its centre at \((0,3)\) is
- A \(6\)
- B \(4\)
- C \(3\)
- D \(2\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
Equation of ellipse, \(\frac{x^2}{16}+\frac{y^2}{9}=1\) \(a^2=16 \Rightarrow a=4, b^2=9 \Rightarrow b=3\) Here, \(a>b\) Now, \(e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}\) \(\therefore \quad \text { Focus }=( \pm a e, 0)= \pm \sqrt{7}, 0\) Since, the…
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