a bracket of 40 B (of minus 40 are two given points the locus of peace
Answers
Answer:
The equation of locus is \frac { x^{ 2 } }{ 4 } -\frac { y^{ 2 } }{ 12 } = 1
4
x
2
−
12
y
2
=1
Step-by-step explanation:
Given:
A = (4,0) and B(-4,0).
Let P(x, y) be the moving points.
Given, |PA-PB|=4
Using distance formula, \sqrt { ((x_{ 1 }-x_{ 2 })^{ 2 }+(y_{ 1 }-y_{ 2 })^{ 2 }) }
((x
1
−x
2
)
2
+(y
1
−y
2
)
2
)
⇒\sqrt { (x-4)^{ 2 }+(y-0)^{ 2 }) } -\sqrt { ((x+4)^{ 2 }+(y-0)^{ 2 }) } =\quad 4
(x−4)
2
+(y−0)
2
)
−
((x+4)
2
+(y−0)
2
)
=4
⇒\sqrt { ((x-4)^{ 2 }+(y)^{ 2 }) } =4+\sqrt { ((x+4)^{ 2 }+(y)^{ 2 }) }
((x−4)
2
+(y)
2
)
=4+
((x+4)
2
+(y)
2
)
Squaring both sides, we get (x-4)^{ 2 }-y^{ 2 }=16+(x+4)^{ 2 }+y^{ 2 }+2×4×\sqrt { ((x+4)^{ 2 }+(y)^{ 2 }) }(x−4)
2
−y
2
=16+(x+4)
2
+y
2
+2×4×
((x+4)
2
+(y)
2
)
⇒x^{ 2 }+16-8x+y^{ 2 }=16+x^{ 2 }+16+8x+y^{ 2 }+2×4×\sqrt { ((x+4)^{ 2 }+(y)^{ 2 }) }x
2
+16−8x+y
2
=16+x
2
+16+8x+y
2
+2×4×
((x+4)
2
+(y)
2
)
⇒-8x=16+8x+8×\sqrt { ((x+4)^{ 2 }+(y)^{ 2 }) }−8x=16+8x+8×
((x+4)
2
+(y)
2
)
⇒-8x-8x-16=8 ×\sqrt { ((x+4)^{ 2 }+(y)^{ 2 }) }−8x−8x−16=8×
((x+4)
2
+(y)
2
)
⇒-16x-16=8 × \sqrt { ((x+4)^{ 2 }+(y)^{ 2 }) }−16x−16=8×
((x+4)
2
+(y)
2
)
⇒-2(x+1) = \sqrt { ((x+4)^{ 2 }+(y)^{ 2 }) }−2(x+1)=
((x+4)
2
+(y)
2
)
Again, squaring both sides, we get ⇒4(x^2+2x+1) = (x+4)^2+(y)^24(x
2
+2x+1)=(x+4)
2
+(y)
2
⇒4x^2+8x+4=x^2+8x+16+y^24x
2
+8x+4=x
2
+8x+16+y
2
⇒3x^2-y^2=123x
2
−y
2
=12
⇒\frac { x^{ 2 } }{ 4 } -\frac { y^{ 2 } }{ 12 } = 1
4
x
2
−
12
y
2
=1 is the equation of locus of the given points.
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