COMEDK · Maths · 14. Ellipse
If an ellipse has an equation in the standard form and it passes through the points \(\left(\dfrac{5}{2}, \dfrac{\sqrt{6}}{4}\right)\) and \(\left(-2, \dfrac{\sqrt{15}}{5}\right)\) then the length of its latus rectum is
- A \(\dfrac{\sqrt{10}}{5}\)
- B \(\sqrt{\dfrac{10}{5}}\)
- C \(\dfrac{1}{\sqrt{10}}\)
- D \(\dfrac{1}{10}\)
Answer & Solution
Correct Answer
(A) \(\dfrac{\sqrt{10}}{5}\)
Step-by-step Solution
Detailed explanation
Let the equation of the ellipse be \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\). Substituting the point \(\left(\dfrac{5}{2}, \dfrac{\sqrt{6}}{4}\right)\) into the equation: \(\dfrac{25}{4a^2} + \dfrac{6}{16b^2} = 1 \Rightarrow \dfrac{25}{4a^2} + \dfrac{3}{8b^2} = 1\).…
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