Find the equation of the set of points P, the sum of whose distance from $A (4, 0, 0)$ and $B (- 4, 0, 0)$ is equal to 10.

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Solution

Let $P (x, y, z)$ be any point.

Then

$P A = \sqrt{(x - 4)^{2} + (y - 0)^{2} + (z - 0)^{2}}$

= $\sqrt{x^{2} + 16 - 8 x + y^{2} + z^{2}}$

$P B = \sqrt{(x + 4)^{2} + (y - 0)^{2} + (z - 0)^{2}}$

= $\sqrt{x^{2} + 16 + 8 x + y^{2} + z^{2}}$

It is given that $P A + P B = 10$

$∴ \sqrt{x^{2} + 16 - 8 x + y^{2} + z^{2}}$

$+ \sqrt{x^{2} + 16 + 8 x + y^{2} + z^{2} = 10}$

$\Rightarrow \sqrt{x^{2} + 16 - 8 x + y^{2} + z^{2}}$

= $10 - \sqrt{x^{2} + 16 + 8 x + y^{2} + z^{2}}$

Squaring both sides, we have

$x^{2} + 16 - 8 x + y^{2} + z^{2}$

= $100 + x^{2} + 16 + 8 x + y^{2} + z^{2}$

$- 20 \sqrt{x^{2} + 16 + 8 x + y^{2} + z^{2}}$

$\Rightarrow 20 \sqrt{x^{2} + 16 + 8 x + y^{2} + z^{2}} = 16 x + 100$

$\Rightarrow 5 \sqrt{x^{2} + 16 + 8 x + y^{2} + z^{2}} = 4 x + 25$

Squaring both sides again, we have

$25 (x^{2} + 16 + 8 x + y^{2} + z^{2}) = 16 x^{2} + 625 + 200 x$

$\Rightarrow 25 x^{2} + 400 + 200 x + 25 y^{2} + 25 z^{2} - 16 x^{2} - 625 - 200 x = 0$

$\Rightarrow 9 x^{2} + 25 y^{2} + 25 z^{2} - 225 = 0$

Thus the required equation is

$9 x^{2} + 25 y^{2} + 25 z^{2} - 225 = 0.$

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[MP PET 1988]

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