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

Question

Lily Anderson · Accepted Answer

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

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

Anna Allen · Answer

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

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.

Cora Alexander · Answer

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

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.!

Lily Anderson · Answer

undefined To find the Cartesian equation for the locus of points $(x, y)$ given $z = x + iy$ and $|z + 3| + |z - 3| = 8$, we'll first express $z$ in terms of $x$ and $y$, then apply the properties of absolute value to simplify the equation.

We have $z = x + iy$, so we can rewrite the given equation using this expression:

$$ |(x + iy) + 3| + |(x + iy) - 3| = 8 $$

Now, let's work with each absolute value term separately:

1. For $|z + 3|$:
$$ |(x + iy) + 3| = |x + iy + 3| $$

2. For $|z - 3|$:
$$ |(x + iy) - 3| = |x + iy - 3| $$

Using the properties of absolute value, we know that $|a + bi| = \sqrt{a^2 + b^2}$ for any complex number $a + bi$. So, let's apply this property:

1. For $|z + 3|$:
$$ |x + iy + 3| = \sqrt{(x + 3)^2 + y^2} $$

2. For $|z - 3|$:
$$ |x + iy - 3| = \sqrt{(x - 3)^2 + y^2} $$

Now, substitute these expressions back into the original equation:

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

This equation represents the locus of points $(x, y)$ in the Cartesian plane that satisfy the given condition. This is the Cartesian equation for the locus of points.