Question

Can you find the cartesian equation for the locus of points `(x, y)` if `z=x+iy` and `|z+3| + |z-3| = 8`?

Answer 1

`7 x^2 + 16 y^2=112`

Explanation:

Considering `z=x+iy` and `absz=sqrt(z bar z)` with `bar z = x -iy` then

`abs(z+3)+abs(z-3)=sqrt((x+iy+3)(x-iy+3))+sqrt((x+iy-3)(x-iy-3))`

or

`abs(z+3)+abs(z-3)=sqrt((x+3)^2+y^2)+sqrt((x-3)^2+y^2)`

then with

`sqrt((x+3)^2+y^2)+sqrt((x-3)^2+y^2)=8`

squaring

`(x+3)^2+y^2 =8^2-16sqrt((x-3)^2+y^2)+(x-3)^2+y^2`

or

`6x=8^2-16sqrt((x-3)^2+y^2)-6x` or

`16sqrt((x-3)^2+y^2)=8^2-12x` or

`4sqrt((x-3)^2+y^2)=16-3x` sqaring again

`16((x-3)^2+y^2)=(16-3x)^2` giving

`7 x^2 + 16 y^2=112` which is the equation of an ellipse

Answer 2

The locus of the points is an ellipse `x^2/16+y^2/7=1`

Explanation:

The modulus of a complex number `a+ib`, is

`|a+ib|=sqrt(a^2+b^2)`

Therefore,

`|z+3|+|z-3|=8`

As, `z=x+iy`

`|x+iy+3|+|x+iy-3|=8`

`sqrt((x+3)^2+y^2)+sqrt((x-3)^2+y^2)=8`

`sqrt((x+3)^2+y^2)=8-sqrt((x-3)^2+y^2)8`

Squaring both sides

`(sqrt((x+3)^2+y^2))^2=(8-sqrt((x-3)^2+y^2))^2`

`(x+3)^2+y^2=64-16sqrt((x-3)^2+y^2)+(x-3)^2+y^2`

`x^2+6x+9=64-16sqrt((x-3)^2+y^2)+x^2-6x+9`

`12x-64=-16sqrt((x-3)^2+y^2)`

`3x-16=-4sqrt((x-3)^2+y^2)`

Squaring both sides

`(3x-16)^2=16((x-3)^2+y^2)`

`9x^2-96x+256=16(x^2-6x+9+y^2)`

`9x^2-96x+256=16x^2-96x+504+16y^2`

`9x^2+256=16x^2+16y^2+144`

`7x^2+16y^2=112`

`7/112x^2+16/112y^2=1`

`x^2/16+y^2/7=1`

This is the equation of an ellipse.

Answer 3

The reqd. Locus `={(x,y) : 7x^2+16y^2=112} sub RR^2.`

Explanation:

Respected Cesareo R. Sir has solved the Problem using

Algebraic Method. We solve it with the help Geometry.

Let `P(x,y)` denote the complex no. `z=x+iy,` and, let

`S(3,0) & S'(-3,0)` be the fixed pts. of the Plane.

With these, note that, `|z-3| and |z+3|` denotes the Distances

`SP and S'P,` resp.

Now, by what is given,

`SP+S'P=8=2a," say, giving, "a=4......................(ast).`

Now, we know from Geometry that, under the condition `(ast),`

the variable pt.`P(x,y)` describes an Ellipse given by,

`x^2/a^2+y^2/b^2=1` having Focii `S(ae,0), and, S'(-ae,0)` and

its Eccentricity `e<1," given by, "b^2=a^2(1-e^2).`

Here, Length of Major Axis is `2a`, & that of Minor, `2b.`

So, `ae=3, a=4 rArr e=3/4 rArr b^2=16(1-9/16)=7.`

Hence, the Ellipse is `x^2/16+y^2/7=1, i.e., 7x^2+16y^2=112.`

Thus, the reqd. Locus `={(x,y) : 7x^2+16y^2=112} sub RR^2.`

Enjoy Maths.!