i done that it will be

sqrt[ (x-1)^2 + y^2 ] + sqrt[ (x+1)^2 + y^2 ] =2

bt how do we solve this?

Hi sandy,

\(\displaystyle \sqrt{(x-1)^2+y^2}+\sqrt{(x+1)^2+y^2}=2\)

\(\displaystyle \sqrt{(1+x)^2+y^2}=2-\sqrt{(1-x)^2+y^2}\)

Square both sides

\(\displaystyle \sqrt{(1+x)^2+y^2}\sqrt{(1+x)^2+y^2}=\left(2-\sqrt{(1-x)^2+y^2}\right)\left(2-\sqrt{(1-x)^2+y^2}\right)\)

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

\(\displaystyle y^2\) is common

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

\(\displaystyle 1+x^2\) is common

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

\(\displaystyle 4x=4\left(1-\sqrt{(1-x)^2+y^2}\right)\)

\(\displaystyle x=1-\sqrt{(1-x)^2+y^2}\)

\(\displaystyle 1-x=\sqrt{(1-x)^2+y^2}\)

Square both sides

\(\displaystyle (1-x)^2=(1-x)^2+y^2\)

The only way these can be equal is if y=0.

You should note that you can see the solution geometrically,

by just following mr fantastic's hint.

The distance from (-1,0) to (1,0) is 1-(-1)=2.

The shortest distance between 2 points is a straight line.

Hence any point

__not__ between (-1,0) and (1,0) must be a combined distance >2 from those two points.

Any point between (-1,0) and (1,0) is a combined distance of 2 from the two points.

Visually, placing any point above the line segment, you can see that the

combined hypotenuses of the two right-angled triangles formed

have sum of lengths > the combined lengths of the bases.

Hence, any point not on the line segment is more than 2 combined units from those points.