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