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NDA 2020 Maths · Mathematics · Conic Sections · easy

The sum of the distances of any point on the ellipse 16x^2+25y^2=400 from its two foci is

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The sum of the distances of any point on the ellipse 16x2+25y2=40016x^2+25y^2=400 from its two foci is
  1. A. 55
  2. B. 88
  3. C. 66
  4. D. 1010
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Answer: (D) 1010

Divide by 400400: x225+y216=1\dfrac{x^2}{25}+\dfrac{y^2}{16}=1, so a=5a=5 and b=4b=4. For an ellipse the sum of focal distances is the constant 2a=102a=10. The value 55 is the semi-major axis, 88 is 2b2b (the minor axis) and 66 is the distance 2c2c between the foci, since c=25−16=3c=\sqrt{25-16}=3. Do not confuse these related lengths.
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Original question written in the NDA 2020 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake