Describe locus of points $z$ that satisfy $|z+2|+|z-2|=5$

$\begingroup$

For this problem, as the question says, I am supposed to describe the locus of points $z$ that satisfy the equation:

$$|z+2|+|z-2|=5$$

Usually these problems aren't too difficult with a bit of algebra, but this one is just confusing me too much (it's getting way too messy).

When I rewrite the equation in terms of $z$'s $x$ and $y$ components, I get:

$$\sqrt{(x+2)^2 + y^2} + \sqrt{(x-2)^2 + y^2} = 5$$

From here, if I choose to square the entire equation, I will get a nasty term in a radical that's too much for me to work out. If I multiply the entire equation by the conjugate (not complex conjugate) of the left-hand side, I arrive at nearly the exact same problem on the right side.

Please give me some guidance. Is the algebra avoidable?

$\endgroup$ 3

3 Answers

$\begingroup$

Let's try some messy algebra. You arrived at:

$$ \sqrt{(x+2)^{2}+y^{2}} = 5-\sqrt{(x-2)^{2}+y^{2}} $$

Squaring both sides, we get:

$$ (x+2)^{2}+y^{2}=25-10\sqrt{(x-2)^{2}+y^{2}} + (x-2)^{2}+y^{2} \to 8x=25-10\sqrt{(x-2)^{2}+y^{2}} $$

We move the $25$ to the LHS and square it again:

$$ (8x-25)^{2}=100((x-2)^{2}+y^{2}) \to \\ 64x^{2}-400x+625=100x^{2}-400x+400+100y^{2} \to \\ 100y^{2}=-36x^{2}+225 $$

Then, we have as possible for $y$:

$$ y=\pm\frac{\sqrt{-36x^{2}+225}}{10} $$

We note that the expression inside the square root is $\geq 0$ only for $\frac{-5}{2}\leq x\leq \frac{5}{2}$. Therefore, the points that satisfy the condition are given by

$$ \{(x,y)\in \mathbb{R^{2}} | \frac{-5}{2}\leq x\leq \frac{5}{2}, y=\pm\frac{\sqrt{-36x^{2}+225}}{10} \} $$

The positive and negative signs on $y$ give you the upper and lower branches of the ellipse. Here is a link for a plot of the upper branch. Another way (and nicer) to see this is taking the equation involving the squares of $x$ and $y$ and putting it on this form:

$$ 36x^2 + 100y^2 = 225 $$

You can then factor this into

$$ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, $$

where $a=5/2$ and $b=3/2$, which gives you geometric information about the ellipse much faster.

$\endgroup$ 2 $\begingroup$

$$|z+2|+|z-2|=5$$ This is an ellipse in the complex plane.
If you want to see it with in more details, choose $C=0,$ $$F_1=-2, F_2=2$$ and $$a=\frac{5}{2}.$$

$\endgroup$ 1 $\begingroup$

Thinking of it conceptually, it says that the distance from z to -2 plus the distance from 2 is 5, a constant. This makes an ellipse. An ellipse is formed by taking two fixed points (foci) and taking the set whose sum to these two points is a constant.

So it's an ellipse with foci at 2, -2.

There are a lot of shapes you can make by making subtle changes to this equation.

If you try |z+2|-|z-2|=5, you'll get a hyperbola with the same foci.

You can do |z+2||z-2|=b for non-negative b would give you Ovals of Cassini.

|z+2|=b for non-negative b is a circle centered at -2 of radius b.

Edit: For Ovals of Cassini, it is the product, not the ratio of the distances to two fixed points (I have made the change in the post). Also, I specified what b was and added a detail for the circle |z+2|=b example.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy