If $(c, 0)$ and $(-c, 0)$ are the foci of an ellipse, and the sum of the distance of any point on the ellipse with the foci is $2a$ I am asked to prove thath the equation of the ellipse is: $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $$ where $b^2=a^2-c^2$.
I tried to first write the definition of the ellipse in mathematical terms: $$ \sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a $$ But when I tried to proceed from here I didn't know where should I go.
On
Hint:
square the two members to get a form like $p+2\sqrt{pq}+q=4a^2$;
isolate the square root in a member and square again to get a polynomial equation;
simplify and rearrange.
Addendum:
The squaring of $4a^2-p-q$ generates the quartic terms $p^2+2pq+q^2$. After subtraction of $4pq$ from the other member, this makes $(p-q)^2$, allowing the polynomial to remain quadratic.
On
Starting of your definition in the question if $F_1(-c,0)$ and $F_2(c,0)$ are the foci ($c>0$) and $P\in\mathcal E$ with $P(x,y)$ and $O=(0,0)$ you have
$$\mathcal E: \quad |PF_1|+|PF_2|=2a$$
Hence:
$$(x+c)^2+y^2+(x-c)^2+y^2+2\sqrt{[(x+c)^2+y^2][(x-c)^2+y^2]}=4a^2$$
$$2x^2+2c^2+2y^2+2\sqrt{[(x+c)^2+y^2][(x-c)^2+y^2]} = 4a^2$$
$$x^2+c^2+y^2+\sqrt{[(x+c)^2+y^2][(x-c)^2+y^2]} = 2a^2$$
Isolate the root and square it again....
$$\sqrt{[(x+c)^2+y^2][(x-c)^2+y^2]} = 2a^2-x^2-y^2-c^2$$
$$(x^2+c^2+2xc+y^2) (x^2+c^2-2xc+y^2) = 4a^4+x^4+y^4+c^4-4a^2x^2-4a^2y^2-4a^2c^2+2x^2y^2+2x^2c^2+2y^2c^2$$
$$x^4+c^4+y^4+2x^2c^2+2x^2y^2+2c^2y^2-4x^2c^2 =4a^4+x^4+y^4+c^4-4a^2x^2-4a^2y^2-4a^2c^2+2x^2y^2+2x^2c^2+2y^2c^2$$
From which, simplifying,
$$4a^4-4a^2x^2-4a^2y^2-4a^2c^2+4x^2c^2 = 0$$
$$a^4-a^2x^2-a^2y^2-a^2c^2+x^2c^2 = 0$$
$$x^2(c^2-a^2)-a^2y^2 = a^2c^2 -a^4$$
Finally, remembering that $a^2-c^2 \gt 0$ we can put $b^2=a^2-c^2$ and have like this
$$-b^2x^2-a^2y^2=-b^2a^2$$ hence the proof
$$\begin{equation} \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \end{equation}$$
A slightly faster way
Hint:
Move $\sqrt{(x+c)^2+y^2}$ to other side of equality to give $\sqrt{(x-c)^2+y^2}=2a-\sqrt{(x+c)^2+y^2}$
Square both sides.
The calculation should simplify a lot as a lot of terms cancel after first step ,compared to simply squaring original equation.
Then move terms within roots to another side and square again.
The problem should be solved.