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AP EAMCET · Maths · Three Dimensional Geometry

For \(a \neq 0\), if the sum of the distances of a point from the points \((a, 0,0)\) and \((-a, 0,0)\) is a constant \(2 \mathrm{k}\), then the locus of that point is

  1. A \(x^2+k^2\left(y^2+z^2\right)=k^2\)
  2. B \(\frac{x^2}{k^2}+\frac{y^2+z^2}{k^2-a^2}=1\)
  3. C \(\frac{x^2}{k^2-a^2}+\frac{y^2+z^2}{k^2}=1\)
  4. D \(x^2+y^2+z^2=\frac{1}{k^2+1}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{x^2}{k^2}+\frac{y^2+z^2}{k^2-a^2}=1\)

Step-by-step Solution

Detailed explanation

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