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

Mar 15, 2017

$7 {x}^{2} + 16 {y}^{2} = 112$

#### Explanation:

Considering $z = x + i y$ and $\left\mid z \right\mid = \sqrt{z \overline{z}}$ with $\overline{z} = x - i y$ then

$\left\mid z + 3 \right\mid + \left\mid z - 3 \right\mid = \sqrt{\left(x + i y + 3\right) \left(x - i y + 3\right)} + \sqrt{\left(x + i y - 3\right) \left(x - i y - 3\right)}$

or

$\left\mid z + 3 \right\mid + \left\mid z - 3 \right\mid = \sqrt{{\left(x + 3\right)}^{2} + {y}^{2}} + \sqrt{{\left(x - 3\right)}^{2} + {y}^{2}}$

then with

$\sqrt{{\left(x + 3\right)}^{2} + {y}^{2}} + \sqrt{{\left(x - 3\right)}^{2} + {y}^{2}} = 8$

squaring

${\left(x + 3\right)}^{2} + {y}^{2} = {8}^{2} - 16 \sqrt{{\left(x - 3\right)}^{2} + {y}^{2}} + {\left(x - 3\right)}^{2} + {y}^{2}$

or

$6 x = {8}^{2} - 16 \sqrt{{\left(x - 3\right)}^{2} + {y}^{2}} - 6 x$ or

$16 \sqrt{{\left(x - 3\right)}^{2} + {y}^{2}} = {8}^{2} - 12 x$ or

$4 \sqrt{{\left(x - 3\right)}^{2} + {y}^{2}} = 16 - 3 x$ sqaring again

$16 \left({\left(x - 3\right)}^{2} + {y}^{2}\right) = {\left(16 - 3 x\right)}^{2}$ giving

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

Mar 15, 2017

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

#### Explanation:

The modulus of a complex number $a + i b$, is

$| a + i b | = \sqrt{{a}^{2} + {b}^{2}}$

Therefore,

$| z + 3 | + | z - 3 | = 8$

As, $z = x + i y$

$| x + i y + 3 | + | x + i y - 3 | = 8$

$\sqrt{{\left(x + 3\right)}^{2} + {y}^{2}} + \sqrt{{\left(x - 3\right)}^{2} + {y}^{2}} = 8$

$\sqrt{{\left(x + 3\right)}^{2} + {y}^{2}} = 8 - \sqrt{{\left(x - 3\right)}^{2} + {y}^{2}} 8$

Squaring both sides

${\left(\sqrt{{\left(x + 3\right)}^{2} + {y}^{2}}\right)}^{2} = {\left(8 - \sqrt{{\left(x - 3\right)}^{2} + {y}^{2}}\right)}^{2}$

${\left(x + 3\right)}^{2} + {y}^{2} = 64 - 16 \sqrt{{\left(x - 3\right)}^{2} + {y}^{2}} + {\left(x - 3\right)}^{2} + {y}^{2}$

${x}^{2} + 6 x + 9 = 64 - 16 \sqrt{{\left(x - 3\right)}^{2} + {y}^{2}} + {x}^{2} - 6 x + 9$

$12 x - 64 = - 16 \sqrt{{\left(x - 3\right)}^{2} + {y}^{2}}$

$3 x - 16 = - 4 \sqrt{{\left(x - 3\right)}^{2} + {y}^{2}}$

Squaring both sides

${\left(3 x - 16\right)}^{2} = 16 \left({\left(x - 3\right)}^{2} + {y}^{2}\right)$

$9 {x}^{2} - 96 x + 256 = 16 \left({x}^{2} - 6 x + 9 + {y}^{2}\right)$

$9 {x}^{2} - 96 x + 256 = 16 {x}^{2} - 96 x + 504 + 16 {y}^{2}$

$9 {x}^{2} + 256 = 16 {x}^{2} + 16 {y}^{2} + 144$

$7 {x}^{2} + 16 {y}^{2} = 112$

$\frac{7}{112} {x}^{2} + \frac{16}{112} {y}^{2} = 1$

${x}^{2} / 16 + {y}^{2} / 7 = 1$

This is the equation of an ellipse.

Mar 15, 2017

The reqd. Locus $= \left\{\left(x , y\right) : 7 {x}^{2} + 16 {y}^{2} = 112\right\} \subset {\mathbb{R}}^{2.}$

#### Explanation:

Respected Cesareo R. Sir has solved the Problem using

Algebraic Method. We solve it with the help Geometry.

Let $P \left(x , y\right)$ denote the complex no. $z = x + i y ,$ and, let

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

With these, note that, $| z - 3 | \mathmr{and} | z + 3 |$ denotes the Distances

$S P \mathmr{and} S ' P ,$ resp.

Now, by what is given,

$S P + S ' P = 8 = 2 a , \text{ say, giving, } a = 4. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(\ast\right) .$

Now, we know from Geometry that, under the condition $\left(\ast\right) ,$

the variable pt.$P \left(x , y\right)$ describes an Ellipse given by,

${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$ having Focii $S \left(a e , 0\right) , \mathmr{and} , S ' \left(- a e , 0\right)$ and

its Eccentricity $e < 1 , \text{ given by, } {b}^{2} = {a}^{2} \left(1 - {e}^{2}\right) .$

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

So, $a e = 3 , a = 4 \Rightarrow e = \frac{3}{4} \Rightarrow {b}^{2} = 16 \left(1 - \frac{9}{16}\right) = 7.$

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

Thus, the reqd. Locus $= \left\{\left(x , y\right) : 7 {x}^{2} + 16 {y}^{2} = 112\right\} \subset {\mathbb{R}}^{2.}$

Enjoy Maths.!