Question

Find the equation of the locus of a point the difference of whose distance from A(-5,0) and B(5,0) is 8?

Answer 1

Answer:

The locus of P is $9x^2-16y^2=144$

Step-by-step explanation:

Formula used:

The distance between two points $(x_1,y_1)\:and\:(x_2,y_2)\:is\:d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Given points are A(-5,0) and B(5,0)

Let the moving point be P(h, k)

As per given data,

$PA-PB=8$

$\sqrt{(h+5)^2+(k-0)^2}-\sqrt{(h-5)^2+(k-0)^2}=8$

$\sqrt{(h+5)^2+k^2}=8+\sqrt{(h-5)^2+k^2}$

squaring on both sisdes we get

$(\sqrt{(h+5)^2+k^2})^2=(8+\sqrt{(h-5)^2+k^2})^2$

$(h+5)^2+k^2=64+(h-5)^2+k^2+2*8*\sqrt{(h-5)^2+k^2}$

$h^2+25+10h=64+h^2+25-10h+16\sqrt{(h-5)^2+k^2}$

$10h=64-10h+16\sqrt{(h-5)^2+k^2}$

Rearranging terms we get

$20h-64=16\sqrt{(h-5)^2+k^2}$

$4(5h-16)=16\sqrt{(h-5)^2+k^2}$

$(5h-16)=4\sqrt{(h-5)^2+k^2}$

squaring once again

$(5h-16)^2=16[(h-5)^2+k^2]$

$25h^2+256-160h=16[h^2+25-10h+k^2]$

$9h^2-16k^2=144$

The locus of P is

$9x^2-16y^2=144$

Answer 2

Answer:

9^2x-16^2y=144

Step-by-step explanation: