Question

find the equation of the locus of p, if A=(2,3),B=(2,-3) and PA+PB=8

Answer 1

here is ur answer.......

Answer 2

Answer:

The required equation of the locus is $16x^2+7y^2-64x-48=0$

Step-by-step explanation:

Given :   A=(2,3),B=(2,-3) and PA+PB=8

To find : The equation of the locus of P?

Solution :

A = (2,3) and B(2,-3).  

Let P(x, y) be the moving points.  

Given, PA+PB=8

Using distance formula,

$D=\sqrt {(x_{ 1 }-x_{ 2 })^{ 2 }+(y_{ 1 }-y_{ 2 })^{ 2 }}$

$PA=\sqrt{(2-x)^2+(3-y)^2}$

$PB=\sqrt{(2-x)^2+(-3-y)^2}$

Substitute,

$PA=8-PB$

$\sqrt{(2-x)^2+(3-y)^2}=8-\sqrt{(2-x)^2+(-3-y)^2}$

Squaring both side,

$(2-x)^2+(3-y)^2=64+(2-x)^2+(-3-y)^2-16(\sqrt{(2-x)^2+(-3-y)^2})$

$9+y^2-6y=64+9+y^2+6y-16(\sqrt{(2-x)^2+(-3-y)^2})$

$-12y=64-16(\sqrt{(2-x)^2+(-3-y)^2})$

$64+12y=16(\sqrt{(2-x)^2+(-3-y)^2})$

$16+3y=4(\sqrt{(2-x)^2+(-3-y)^2})$

Again Squaring both side,

$(16+3y)^2=16((2-x)^2+(-3-y)^2))$

$256+9y^2+96y=16(4+x^2-4x+9+y^2+6y)$

$256+9y^2+96y=208+16x^2-64x+16y^2+96y$

$16x^2+7y^2-64x-48=0$

Therefore, The required equation of the locus is $16x^2+7y^2-64x-48=0$