<s>[INST] locus of points with sum of distance 
Find the quadratic equation for the locus of points whose sum of its distance from $(1, 0)$ and $(−1, 0)$ is $6$.

I know to start with $\sqrt{(x-1)^2 +y^2} + \sqrt{(x+1)^2+y^2}=6$, then square both sides. I'm not sure about the algebra after squaring both sides as I got that part wrong. 
 [/INST] $$\sqrt{(x-1)^2 +y^2} + \sqrt{(x+1)^2+y^2}=6$$
$$\sqrt{(x+1)^2+y^2}=6-\sqrt{(x-1)^2+y^2}$$
square both sides ($LHS>0;\;RHS>0$)
$$(x+1)^2+y^2=\left(6-\sqrt{(x-1)^2+y^2}\right)^2$$
$$x^2+2 x+y^2+1=36+x^2-2 x+1+y^2-12 \sqrt{(x-1)^2+y^2}$$
bring every term but the square root in the RHS
$$12 \sqrt{(x-1)^2+y^2}=36-4 x$$
simplify dividing both sides by $4$
$$3 \sqrt{(x-1)^2+y^2}=9-x$$
Notice that both sides are positive, so square again
$$9 \left((x-1)^2+y^2\right)=(9-x)^2$$
expand and bring variables into LHS and constants in the RHS
$$8 x^2+9 y^2=72$$
divide both sides by $72$
$$\frac{x^2}{9}+\frac{y^2}{8}=1$$
Which is the equation of an ellipse center $(0,0)$ and semiaxes $a=3;\;b=2\sqrt{2}$
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