Find the locus of a point, sum of whose distances from $(c, 0)$ and $(-c, 0)$ is constant and greater than $2c$

Question

Find the locus of a point, sum of whose distances from $(c, 0)$ and $(-c, 0)$ is constant and greater than $2c$.
Try
Let the point be $(h, k)$. Then by given condition
$\sqrt {(h-c) ^2+k^2} + \sqrt {(h+c) ^2 +k^2} = 2c + \alpha$, where $2c +\alpha$ is a constant. We consider $2c+\alpha$ because given that the sum is a constant greater than $2c$.So $\alpha \in \Bbb {R^+} $.
After this stage I have squared this equation to make it free of square root but unable to determine the locus. Please help and thanks in advance.

Answer 1

Yeah, basically it s ellipse.. Let $2a=2c+\alpha$ then we can derive the standatd ellipse equation from it. The trick to deal with the sqrt is to square twice.

$$ \sqrt {(x-c) ^2+y^2} + \sqrt {(x+c) ^2 +y^2} = 2c + \alpha := 2a\\ \sqrt {x^2+y^2+c^2-2cx} + \sqrt {x^2+y^2+c^2+2cx} = 2a $$ Let $A=x^2+y^2+c^2$ Square once $$ A-2cx+A+2cx+2\sqrt{A^2-4c^2x^2}=4a^2\\ 2a^2-A=\sqrt{A^2-4c^2x^2}\\ A^2-4a^2A+4a^4=A^2-4c^2x^2 $$ Then you get the ellipse equation $$ a^4=a^2A-c^2x^2\\ a^2(a^2-c^2)=(a^2-c^2)x^2+a^2y^2\\ 1=\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}\\ $$