Question

If z = x+iy and |z-1| + |z+1| = 4, show that 3x² + 4y² = 12
Please help!

Answer 1

HELLO THERE!

z = x + iy

Also,

|z - 1| + |z + 1| = 4

Put z = x + iy in the equation:

|x + iy - 1| + |x + iy + 1| = 4

=> |(x - 1) + iy| + |(x + 1) + iy| = 4

$\implies \sqrt{(x-1)^{2} + y^{2}} + \sqrt{(x+1)^{2} + y^{2}} = 4 \\\\\implies \sqrt{(x-1)^{2} + y^{2}} = 4 - \sqrt{(x+1)^{2} + y^{2}} \\\\\implies (x-1)^{2} + y^{2} = 16 - 8\sqrt{(x+1)^{2} + y^{2}} + (x+1)^{2} + y^{2} [S.B.S] \\\\\implies x^{2} - 2x + 1 + y^{2} = 16 - 8\sqrt{(x+1)^{2} + y^{2}} + x^{2} + 2x + 1 + y^{2} \\\\\implies 16 - 8\sqrt{(x+1)^{2} + y^{2}} + 4x = 0 \\\\\implies 8\sqrt{(x+1)^{2} + y^{2}} = 16 + 4x \\\\\implies 2\sqrt{(x+1)^{2} + y^{2}} = 4 + x$

$\\\\\implies 4[(x+1)^{2} + y^{2}] = 16 + 8x + x^{2} \\\\\implies 4[x^{2} + 2x + 1 + y^{2}] = x^{2} + 8x + 16 \\\\\implies 4x^{2} + 8x + 4 + 4y^{2} = x^{2} + 8x + 16 \\\\\implies 3x^{2} + 4y^{2} = 12$

Hence, your result is proved.

HOPE MY ANSWER IS SATISFACTORY..

Thanks!