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Mathematics · JEE | CONIC SECTIONS

हिन्दी

Find the equation of the ellipse whose vertices are \( (\pm 13, 0) \) and foci are \( (\pm 5, 0) \).

  1. A
    \( \frac{x^2}{169} + \frac{y^2}{25} = 1 \)
  2. B
    \( \frac{x^2}{144} + \frac{y^2}{169} = 1 \)
  3. C
    \( \frac{x^2}{25} + \frac{y^2}{169} = 1 \)
  4. D
    \( \frac{x^2}{169} + \frac{y^2}{144} = 1 \)
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 \( \frac{x^2}{169} + \frac{y^2}{144} = 1 \).
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