# Solving equations involving square roots

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 2 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 2 at 14:59
• All possible solutions – Shantanu Kaushik Feb 2 at 15:02
• i might be wrong, but that is known as ellipse – aaaaaa Feb 2 at 18:18

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 &\\ \end{align} and that \begin{align}(x^2 +(y-2)^2) - (x^2+(y+2)^2) &= -8y &\\ \end{align}

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

Adding $$$$ and $$$$ gives us \begin{align} 2\sqrt{x^2 +(y-2)^2} &= 6-\frac{4}{3}y &\\ 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}

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: $$\frac{x^2}{5}+\frac{y^2}{9} = 1$$

• 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 2 at 15:13
• You can simply edit your answer. – Yves Daoust Feb 2 at 15:17
• Thank you kind sir – Shantanu Kaushik Feb 2 at 15:33

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

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

$$1296-144(x^2+y^2+4)+64y^2=0$$
$$\left(\frac x{\sqrt5}\right)^2+\left(\frac y3\right)^2=1,$$ which is a centered, axis-aligned ellipse.
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: $$x^2+(y-2)^2=\left(\frac{9-2y}{3}\right)^2 \Rightarrow \\ 9x^2+9y^2-36y+36=81-36y+4y^2 \Rightarrow \\ 9x^2+5y^2=45 \Rightarrow \\ \frac{x^2}{5}+\frac{y^2}{9}=1.$$