Solving Equation of the form $\sqrt{(a+\frac{b}{2})^2+L^2}-\sqrt{(a-\frac{b}{2})^2+L^2}=c$

Question

I have been struggling to find the solution to one of my physics problems mathematically as this is the equation I arrive at where all of the values are known except $a$. I have tried solving for a but I'm just not sure how to isolate it. $\sqrt{(a+\frac{b}{2})^2+L^2}-\sqrt{(a-\frac{b}{2})^2+L^2}=c$

Answer 1

Consider the hyperbola defined by this equation:$$\sqrt{\left(x+\frac{b}{2}\right)^2+y^2}-\sqrt{\left(x-\frac{b}{2}\right)^2+y^2}=c$$ The length of the transverse axis will be $c$ and the focal distance will be $b$. From this we can calculate the length of the conjugate axis to be $\sqrt{b^2-c^2}$.

Now we know that this hyperbola can be written in standard form like this:$$\frac{4x^2}{c^2}-\frac{4y^2}{b^2-c^2}=1$$ So for your case you can now put $x=a$ and $y=L$ and easily obtain the solution. Note that no solution will exist if $b<c$

Answer 2

$\sqrt A-\sqrt B~=~C\qquad=>\qquad\sqrt A~=~C+\sqrt B\quad=>\quad A=C^2+B+2C\sqrt B\quad=>$

$A-B-C^2=2C\sqrt B\quad=>\quad\Big(A-B-C^2\Big)^2=4BC^2.~$ This is a polynomial expression,

containing no radicals.