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NDA 2024 Maths · Mathematics · Conic Sections · hard

The locus of a point whose distances from (3,0) and (-3,0) add up to 10 is

Single correct option

The locus of a point whose distances from (3,0)(3,0) and (−3,0)(-3,0) add up to 1010 is
  1. A. x225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1
  2. B. x216+y225=1\frac{x^2}{16}+\frac{y^2}{25}=1
  3. C. x2100+y291=1\frac{x^2}{100}+\frac{y^2}{91}=1
  4. D. x225+y29=1\frac{x^2}{25}+\frac{y^2}{9}=1
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Answer: (A) x225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1

The locus is an ellipse with foci (±3,0)(\pm3,0), so c=3c=3, and the constant sum equals 2a=102a=10, so a=5a=5. Then b2=a2−c2=25−9=16b^2=a^2-c^2=25-9=16. Trap: taking a=10a=10 gives the second option, and b2=9b^2=9 confuses c2c^2 with b2b^2.
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