Solving equations involving square roots

Question

I am a student and I often encounter these type of equations:

$$\sqrt{x^2 + (y-2)^2} + \sqrt{x^2 + (y+2)^2} = 6$$

I usually solve these by taking one term ($\sqrt{x^2 + (y-2)^2}$ for example) to the right hand side but this seems to take more time. Please suggest me some methods which can help me solve these types of problem quickly.

Thanks

What do you want to solve for? For a variable, or in general a parametric form for all possible solutions? — KKZiomek, Feb 02 '19 at 14:56
I wanted to find the locus of a point such that the sum of the distance of that point from (0,2) and (0,-2) is 6 — Shantanu Kaushik, Feb 02 '19 at 14:59

score 8 · Answer 1 · edited Feb 02 '19 at 19:00

8

The left hand side of this equation us equivalent to the distance from the point (x, y) and the point (0, 2) added to the distance from the point (x, y) and the point (0, -2). If these distances add to 6, then we can see that the solution forms an ellipse with centre (0, 0) foci (0, 2), (0, -2). By solving for the points where the distance from both of the foci is 3 we get the vertices at $(\sqrt 5, 0)$, $(-\sqrt5, 0)$, $(0, 3)$, $(0, -3)$. So the equation of the ellipse and hence the solution is given by: $$\frac{x^2}{5}+\frac{y^2}{9} = 1$$

edited Feb 02 '19 at 19:00

S. Sharma

543
2
14

answered Feb 02 '19 at 15:05

Peter Foreman

19,947

I actually made a mistake in writing "for the points where the distance from both of the foci is 3". I meant to say that we can find the vertices by using the fact that they lie where y=0 or x=0 and using Pythagoras' or linear equations in each case. – Peter Foreman Feb 02 '19 at 15:13
1

You can simply edit your answer. – Feb 02 '19 at 15:17
Thank you kind sir – Shantanu Kaushik Feb 02 '19 at 15:33

score 4 · Answer 2 · 2019-02-02T15:14:22.467

4

Use

$$\sqrt a+\sqrt b=c$$ then

$$a+2\sqrt{ab}+b=c^2$$

then

$$4ab=(c^2-a-b)^2=c^4-2c^2(a+b)+(a+b)^2$$

and finally

$$c^4-2c^2(a+b)+(a-b)^2=0.$$

In your case,

$$1296-144(x^2+y^2+4)+64y^2=0$$

or

$$\left(\frac x{\sqrt5}\right)^2+\left(\frac y3\right)^2=1,$$ which is a centered, axis-aligned ellipse.

edited Feb 02 '19 at 15:14

answered Feb 02 '19 at 15:09

Thank you very much sir – Shantanu Kaushik Feb 02 '19 at 15:34

score 3 · Answer 3 · answered Feb 02 '19 at 16:27

Denote: $$\begin{cases}x^2 + (y-2)^2=t^2 \\ x^2+(y+2)^2=t^2+8y\end{cases}.$$ Then: $$\sqrt{x^2 + (y-2)^2} + \sqrt{x^2 + (y+2)^2} = 6 \Rightarrow \\ t+\sqrt{t^2+8y}=6 \Rightarrow \\ t^2+8y=36-12t+t^2 \Rightarrow \\ t=\frac{9-2y}{3}.$$ Plug it into the first equation: $$x^2+(y-2)^2=\left(\frac{9-2y}{3}\right)^2 \Rightarrow \\ 9x^2+9y^2-36y+36=81-36y+4y^2 \Rightarrow \\ 9x^2+5y^2=45 \Rightarrow \\ \frac{x^2}{5}+\frac{y^2}{9}=1.$$

score 2 · Accepted Answer · answered Feb 02 '19 at 23:22

A nice way of doing problems like these is taking advantage of our knowledge of the factorization of difference of squares.

Given that $$\begin{align}\sqrt{x^2 +(y-2)^2} + \sqrt{x^2+(y+2)^2} &= 6 &[1]\\ \end{align}$$ and that $$\begin{align}(x^2 +(y-2)^2) - (x^2+(y+2)^2) &= -8y &[2]\\ \end{align}$$

we have (by $\frac{[2]}{[1]}$) $$\begin{align}\sqrt{x^2 +(y-2)^2} - \sqrt{x^2+(y+2)^2} = \frac{-8y}{6} &= -\frac{4}{3}y &[3]\\ \end{align}$$

Adding $[1]$ and $[3]$ gives us $$\begin{align} 2\sqrt{x^2 +(y-2)^2} &= 6-\frac{4}{3}y &[4]\\ 4(x^2 +(y-2)^2) &= \bigg(6-\frac{4}{3}y\bigg)^2\\ 4x^2 +4y^2-16y+16 &= \frac{16}{9}y^2-16y+36\\ x^2 +y^2+4 &= \frac{4}{9}y^2+9\\ x^2 +\frac{5}{9}y^2 &= 5\\ \frac{x^2}{5} +\frac{y^2}{9} &= 1\\ \end{align}$$

Solving equations involving square roots

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