find the equation of the locus of p A= (2 3) and B=( 2 - 3 ) and PA+PB =8
Answers
Answer:
The required equation of the locus is 16x^2+7y^2-64x-48=016x
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 }}D=
(x
1
−x
2
)
2
+(y
1
−y
2
)
2
PA=\sqrt{(2-x)^2+(3-y)^2}PA=
(2−x)
2
+(3−y)
2
PB=\sqrt{(2-x)^2+(-3-y)^2}PB=
(2−x)
2
+(−3−y)
2
Substitute,
PA=8-PBPA=8−PB
\sqrt{(2-x)^2+(3-y)^2}=8-\sqrt{(2-x)^2+(-3-y)^2}
(2−x)
2
+(3−y)
2
=8−
(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})(2−x)
2
+(3−y)
2
=64+(2−x)
2
+(−3−y)
2
−16(
(2−x)
2
+(−3−y)
2
)
9+y^2-6y=64+9+y^2+6y-16(\sqrt{(2-x)^2+(-3-y)^2})9+y
2
−6y=64+9+y
2
+6y−16(
(2−x)
2
+(−3−y)
2
)
-12y=64-16(\sqrt{(2-x)^2+(-3-y)^2})−12y=64−16(
(2−x)
2
+(−3−y)
2
)
64+12y=16(\sqrt{(2-x)^2+(-3-y)^2})64+12y=16(
(2−x)
2
+(−3−y)
2
)
16+3y=4(\sqrt{(2-x)^2+(-3-y)^2})16+3y=4(
(2−x)
2
+(−3−y)
2
)
Again Squaring both side,
(16+3y)^2=16((2-x)^2+(-3-y)^2))(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=16(4+x
2
−4x+9+y
2
+6y)
256+9y^2+96y=208+16x^2-64x+16y^2+96y256+9y
2
+96y=208+16x
2
−64x+16y
2
+96y
16x^2+7y^2-64x-48=016x
2
+7y
2
−64x−48=0
Therefore, The required equation of the locus is 16x^2+7y^2-64x-48=016x
2
+7y
2
−64x−48=0