A $(4, 0)$ and $B (- 4, 0)$ are two given points. $A$ variable point $P$ is such that $P A + P B = 10 .$ Show that the equation of locus of $P$ is $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$

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Solution

$A (4, 0) B (- 4, 0)$ Let $P (x, y)$

$P A + P B = 10$

$\sqrt{(x - 4)^{2} + y^{2}} + \sqrt{(x + 4)^{2} + y^{2}} = 10$

Squaring, we get

$(x - 4)^{2} + y^{2} + (x + 4)^{2} + y^{2} + 2 \sqrt{[(x - 4)^{2} + y^{2}] [(x + 4)^{2} + y^{2}]} = 100$

$2 y^{2} + 2 (x^{2} + 16) + 2 \sqrt{y^{4} + y^{2} [2 (x^{2} + 16)] + (x^{2} - 16)^{2}} = 100$

$\sqrt{y^{4} + 2 y^{2} (x^{2} + 16) + (x^{2} - 16)^{2}} = 34 - (x^{2} + y^{2})$

Squaring again, we get

$y^{4} + 2 x^{2} y^{2} + 32 y^{2} + x^{2} - 32 x^{2} + 256 = 1156 + x^{4} + y^{4} + 2 x^{2} y^{2} - 68 (x^{2} + y^{2})$

$68 x^{2} - 32 x^{2} + 68 y^{2} + 32 y^{2} = 1156 - 256$

$36 x^{2} + 100 y^{2} = 900$

$98 x^{2} + 25 y^{2} = 225$

( or )

$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ .

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