AP EAMCET · Maths · Ellipse
Let the point L lying in the first quadrant be one end of a latus rectum of the ellipse \(\frac{x^2}{4}+\frac{y^2}{3}=1\). Let \(P\) and \(Q\) be the points where the normal drawn at \(\mathrm{L}\) to this given ellipse meets the major axis and the minor axis. Then the distance between \(\mathrm{P}\) and \(\mathrm{Q}\) is
- A \(\frac{\sqrt{5}}{4}\)
- B \(\frac{1}{\sqrt{2}}\)
- C \(\frac{1}{2 \sqrt{2}}\)
- D \(\frac{\sqrt{5}}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{5}}{4}\)
Step-by-step Solution
Detailed explanation
Given ellipse is \(\frac{x^2}{4}+\frac{y^2}{3}=1\) Here, \(a=2, b=\sqrt{3}\) Now, \(e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{3}{4}}=\frac{1}{2}\) \(\because \mathrm{L}\) is lying in the first quadrant…
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