TS EAMCET · Maths · Ellipse
The length of the latus rectum of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) \((a \gt b)\) is \(\frac{8}{3}\). If the distance from the centre of the ellipse to its focus is \(\sqrt{5}\), then \(\sqrt{a^2+6 a b+b^2}=\)
- A 7
- B \(12 \sqrt{2}\)
- C \(3 \sqrt{5}\)
- D 11
Answer & Solution
Correct Answer
(A) 7
Step-by-step Solution
Detailed explanation
\begin{aligned} & a c=\sqrt{5}...(i) \\ & e=\sqrt{1-\frac{b^2}{a^2}} \Rightarrow e^2=1-\frac{b^2}{a^2} \\ & \frac{5}{a^2}=1-\frac{b^2}{a^2} \Rightarrow 5=a^2-b^2 \text {.[From (i)] } \\ & \Rightarrow b^2=a^2-5....(ii) \\ & \frac{2 b^2}{a}=\frac{8}{3} \Rightarrow…
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