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…

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. $x^2+k^2\left(y^2+z^2\right)=k^2$
  2. $\frac{x^2}{k^2}+\frac{y^2+z^2}{k^2-a^2}=1$
  3. $\frac{x^2}{k^2-a^2}+\frac{y^2+z^2}{k^2}=1$
  4. $x^2+y^2+z^2=\frac{1}{k^2+1}$

Solution

No solution. Refer to answer key.

Asked in: AP EAMCET 2018 (24 Apr Shift 2)

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