Question

Find the cartesian equation of the locus of z in the complex plane satisfying |z-4| + |z+4| = 16

Answer 1

Answer:

$\frac{x^2}{64}+\frac{y^2}{48}=1$

Step-by-step explanation:

|z-4| + |z+4| = 16

Let z=x+iy

Then

|(x+iy)-4|+|(x+iy)+4|=16

|(x-4)+iy|+|(x+4)+iy|=16

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

squaring on both sides

$(x-4)^2+y^2=256+(x+4)^2+y^2-32\sqrt{(x+4)^2+y^2}\\(x-4)^2+y^2=256+(x+4)^2+y^2-32\sqrt{(x+4)^2+y^2}$

cancelling like terms on both sides

we get

$(x-4)^2=256+(x+4)^2-32\sqrt{(x+4)^2+y^2}\\x^2+16-8x=256+x^2+16+8x-32\sqrt{(x+4)^2+y^2}$

$-8x=256+8x-32\sqrt{(x+4)^2+y^2}\\32\sqrt{(x+4)^2+y^2}=256+16x\\32\sqrt{(x+4)^2+y^2}=16(x+16)\\2\sqrt{(x+4)^2+y^2}=x+16$

squaring on both sides

$4[(x+4)^2+y^2]=(x+16)^2\\4[x^2+16+8x+y^2]=x^2+256+32x\\4x^2+64+32x+4y^2=x^2+256+32x$

rearranging terms we get

$3x^2+4y^2=192$

$\frac{x^2}{64}+\frac{y^2}{48}=1$

This equation represents an ellipse.

Therefore the locus of z is an ellipse.

Answer 2

Answer:

Step-by-step explanation:

find the answer below