Question

11. Find the equation of the locus of P, A=(2,3) B=(2,-3) and PA+PB=8.
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Answer 1

Answer:

Step-by-step explanation:

Given,

Co-ordinates of :-

A = ( 2 , 3 )

B = ( 2 , - 3 )

Let. Co-ordinated of 'P' be :-

P = (x , y)

PA + PB = 8

To Find :-

Equation of the locus.

Solution :-

PA + PB = 8

PA = 8 - PB

Squaring on both sides :-

$(PA)^2=(8-PB)^2$

$(PA)^2=64-16PB+(PB)^2$

Let it be equation - 1

Distance Formula :-

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

According to PA :-

(x , y) = (x_1 , y_1)

(2 , 3) = (x_2 , y_2)

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

$= 4 - 4x + x^2 + 9 - 6y + y^2$

$= x^2+y^2-4x-6y+13$

According to PB :-

(x , y) = (x_1 , y_1)

(2 , -3) = (x_2 , y_2)

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

$=4-4x+x^2+y^2+6y+9$

$=x^2+y^2-4x+6y+13$

Substituting Value in equation - 1:-

$x^2+y^2-4x-6y+13 = 64-16PB+x^2+y^2-4x+6y+13$

$-12y-64=-16PB$

$-4(3y+16)=-4(4PB)$

$3y+16=4PB$

Squaring on Both sides :-

$(3y+16)^2=(4PB)^2$

$9y^2+96y+256=16(PB)^2$

Substituting Value of (PB)^2 :-

$9y^2+96y+256=16(x^2+y^2-4x+6y+13)$

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

$16x^2+16y^2-64x+96y+208-9y^2-96y-256=0$

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

Since the equation of the locus is :-

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