Question

The equation of a locus of a point, difference of whose distance from $(-5,0)$ and $(5,0)$ is $8$ unit is

• A. $\frac{x^2}{16}+\frac{y^2}{9}=1$
• B. $\frac{x^2}{16}-\frac{y^2}{9}=1$
• C. $x^2+y^2=25$
• D. $x^2-y^2=9$

Answer 1

Answer: (B)

$\frac{x^2}{16}-\frac{y^2}{9}=1$

Let point be $(x,y)$

So,

$\sqrt{{(x+5)}^2+y^2}-\sqrt{{(x-5)}^2+y^2}=\pm 8$

Considering $8$,we get

${(x+5)}^2+y^2=64+{(x-5)}^2+y^2+16\sqrt{{(x-5)}^2+y^2}10\times 2x-64=16\sqrt{{(x-5)}^2+y^2}{(5x-16)}^2=16[{(x-5)}^2+y^2]25x^2+256-160x=16x^2+16y^2-160x+400144=16y^2-9x^2\frac{y^2}{9}-\frac{x^2}{16}=1$

Considering $-8$, we get

${(x+5)}^2+y^2+64+16\sqrt{{(x+5)}^2+y^2}={(x-5)}^2+y^2-20x-64=16\sqrt{{(x+5)}^2+y^2}{(5x+16)}^2=16({(x+5)}^2+y^2)25x^2+256+160x=160x+16x^2+16y^2+4009x^2-16y^2=144\frac{x^2}{16}-\frac{y^2}{9}=1$