Find $y$ in $\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$

Question

I would like to solve this equation for $y$. Any tips? It seems like you cant really do it analytically?

$$\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$$

Answer 1

Consider points $A(2,6)$, $B(-4,3)$ and a point $P(0,y)$ on y-axis. Then notice that given equation is actually $$AP+BP= AB$$ so $P$ is (by triangle inequality ) on a line $AB$ so it is an intersection point of the line $AB$: $$ y= {1\over 2}x+5$$ and $y$-axis, so $y=5$.

Answer 2

Hint: After squaring one times we get $$2\sqrt{4+(y-6)^2}\sqrt{16+(y-3)^2}=25-(y-6)^2-(y-3)^2$$ squaring again and simplfying we get $$4 \left(46 y^2-450 y+975\right)=0$$

Answer 3

Set $y-3=z,$ then square both sides and simplify to get $$z^2-3z-8=-\sqrt{4+(z-3)^2}\sqrt{16+z^2}.$$ Squaring again and simplifying the right hand side gives $$(z^2-3z-8)^2=(z^2-6z+13)(16+z^2).$$ Now let $16+z^2=w,$ then expand and simplify to get $$z^2+16z+64=5w.$$ Substituting back for $w,$ simplifying and factoring gives the quadratic $$(z-2)^2=0.$$ Thus, $z=y-3=2,$ so that if there is a solution, then it has to be $y=5.$ Substituting this in the original equation assures us that this is indeed the only solution.

Answer 4

You can factor out $2$ at the second summand.

$$\sqrt{\underbrace{4+(y-6)^2}_{=5}}+2\cdot \sqrt{\underbrace{4+\left(\frac{y-3}2\right)^2}_{=5}}=\sqrt{5}+2\cdot\sqrt{5}$$

Now we see that the following equations has to be true at the same time.

1. $(y-6)^2=1 \Rightarrow y_1=7,y_2=5$

2. $\left(\frac{y-3}2\right)^2=1\Rightarrow y_1=5,y_2=1$

Thus the solution is $y=...$