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JEE Main 2024 · Mathematics · Conic Sections · easy

The eccentricity of the ellipse 9x^2+25y^2=225 is

Single correct option

The eccentricity of the ellipse 9x2+25y2=2259x^2+25y^2=225 is
  1. A. 35\dfrac{3}{5}
  2. B. 1625\dfrac{16}{25}
  3. C. 45\dfrac{4}{5}
  4. D. 54\dfrac{5}{4}
Show answer and solution

Answer: (C) 45\dfrac{4}{5}

Dividing by 225225: x225+y29=1\dfrac{x^2}{25}+\dfrac{y^2}{9}=1, so a2=25a^2=25, b2=9b^2=9. Then e=1−b2/a2=1−925=45e=\sqrt{1-b^2/a^2}=\sqrt{1-\tfrac{9}{25}}=\tfrac45. Option 35\tfrac35 is b/ab/a, and 1625\tfrac{16}{25} is e2e^2.
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