Mathematics · JEE | CONIC SECTIONS
हिन्दीFind the equation of the ellipse whose vertices are \( (\pm 13, 0) \) and foci are \( (\pm 5, 0) \).
View Answer & Solution
Correct answer · Option D
\( \frac{x^2}{169} + \frac{y^2}{144} = 1 \)
Explanation
Given \( a = 13 \) and \( c = 5 \), then \( b^2 = a^2 - c^2 = 169 - 25 = 144 \). Equation is \( \frac{x^2}{169} + \frac{y^2}{144} = 1 \).