Question

find the equation of locus of p if A(4,0) B(-4,0)| |PA-PB|=4

Answer 1

In the attachments I have answered this problem.

The distance formula

sqrt [(x1-x2)^2 + (y1-y2)^2] is applied to find distance between the given two points .

See the attachment for detailed solution

Answer 2

Answer:

The equation of locus is $\frac { x^{ 2 } }{ 4 } -\frac { y^{ 2 } }{ 12 } = 1$

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

⇒$\sqrt { (x-4)^{ 2 }+(y-0)^{ 2 }) } -\sqrt { ((x+4)^{ 2 }+(y-0)^{ 2 }) } =\quad 4$

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

Squaring both sides, we get $(x-4)^{ 2 }-y^{ 2 }=16+(x+4)^{ 2 }+y^{ 2 }+2×4×\sqrt { ((x+4)^{ 2 }+(y)^{ 2 }) }$

⇒$x^{ 2 }+16-8x+y^{ 2 }=16+x^{ 2 }+16+8x+y^{ 2 }+2×4×\sqrt { ((x+4)^{ 2 }+(y)^{ 2 }) }$

⇒$-8x=16+8x+8×\sqrt { ((x+4)^{ 2 }+(y)^{ 2 }) }$

⇒$-8x-8x-16=8 ×\sqrt { ((x+4)^{ 2 }+(y)^{ 2 }) }$

⇒$-16x-16=8 × \sqrt { ((x+4)^{ 2 }+(y)^{ 2 }) }$

⇒$-2(x+1) = \sqrt { ((x+4)^{ 2 }+(y)^{ 2 }) }$

Again, squaring both sides, we get ⇒$4(x^2+2x+1) = (x+4)^2+(y)^2$

⇒$4x^2+8x+4=x^2+8x+16+y^2$

⇒$3x^2-y^2=12$

⇒$\frac { x^{ 2 } }{ 4 } -\frac { y^{ 2 } }{ 12 } = 1$ is the equation of locus of the given points.