I tried to derive a formula of a horizontal ellipse and that of a hyperbola and I cannot succed. Many alleged proofs online seem to me no proof at all other than somehow helping memorize the equations.
$x^2+y^2=1$ is an equation of a unit circle because $\{\text{points}:x^2+y^2=1\}=\{\text{points}:\text{distance from origin is } 1\}$, which is equivalent to say $x^2+y^2=1\iff \text{distance from origin is }1$. So to my understanding, a formula is an equation of an ellipse (or a hyperbola) iff it is equivalent to the definition of the conic section. But many proofs I have seen so far only proves the implication. And they are no different to say $(x,y)\in \mathbb R^2$ is a formula for an ellipse (or a hyperbola).
To prove the equation of a hyperbola, I started with $\left|[(x+c)^2+y^2]^{\frac{1}{2}}-[(x-c)^2+y^2]^{\frac{1}{2}}\right|=d$ where $d>0$. But no matter what I do, I end up with an expression where I cannot cancel a $\frac{1}{2}$ power because I cannot square both sides. For example, I get $2x^2+2y^2+2c^2-d^2=2\left[(x^2+y^2+c^2+2xc)(x^2+y^2-c^2-2xc)\right]^{\frac{1}{2}}$. I am not able to cancel the $\frac{1}{2}$ on the right because $\left[\left(2x^2+2y^2+2c^2-d^2\right)^2\right]^{\frac{1}{2}}$ might equal $d^2 -2x^2-2y^2-2c^2$ and the converse is not true. The same thing happens with an ellipse.
Can someone please help me?
You have already seen (in the comments) a demonstration that if $$\left\lvert\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}\right\rvert = d$$ with $d > 0$ then $$ \frac{x^2}{(d/2)^2} - \frac{y^2}{c^2-(d/2)^2} = 1. $$ Now we can simply write $d = 2a$ in order to match the definition that you quoted, and this implies that $d/2=a,$ which produces the conventional denominator under $x^2.$
Next, recall that for any point $P=(x,y)$ on the hyperbola and for the foci at $F_1 = (c,0)$ and $F_2=(-c,0),$ where $c > 0,$ the definition of the hyperbola says that $\lvert PF_1 - PF_2\rvert = 2a > 0.$ But the points $P,$ $F_1,$ and $F_2$ also form a triangle (possibly degenerate), and the triangle inequality implies that $\lvert PF_1 - PF_2\rvert \leq F_1F_2 = 2c.$
In the case where $\lvert PF_1 - PF_2\rvert = 2c$ the only possible points on the "hyperbola" are the points on the $x$-axis outside the interval between the foci. This is generally not considered a conic section, so we can exclude this case.
Therefore $2c > 2a > 0,$ and so $c > a > 0$ and $c^2 > a^2,$ so $c^2 - a^2 > 0.$ Since $c^2 - a^2$ is positive, we can take its square root; set $b = \sqrt{c^2 - a^2},$ and then we can write $b^2 = c^2 - a^2 = c^2 - (d/2)^2$ so that the denominator of $y^2$ can be written as $b^2.$
That's one of the directions of implication, namely, that if $(x,y)$ is on a hyperbola with foci at $(c,0)$ and $(-c,0)$ and if the absolute difference of the distances from every such point $(x,y)$ to each of the foci is $d,$ then $$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \tag1$$ where $a = d/2$ and $b^2 = c^2 - a^2.$
Now let's do the other direction. We will assume $(x,y)$ satisfies Equation $(1)$. Without loss of generality we can also assume $a > 0,$ since replacing $a$ with $-a$ does not change the solution set of Equation $(1)$. We then show that $(x,y)$ lies on a hyperbola with foci at $(c,0)$ and $(-c,0)$, where $c = \sqrt{a^2 + b^2}$, and that the absolute difference between the distances from $(x,y)$ to each focus is $2a.$
First, note that $(x,y)$ satisfies Equation $(1)$ if and only if $(-x,y),$ $(x,-y),$ and $(-x,-y)$ all satisfy Equation $(1).$ That is, whatever figure is formed by the points that satisfy this equation, the figure is mirror-image symmetric across the $x$ axis and across the $y$ axis. We can therefore describe the entire figure by finding the part of it in the right half-plane, that is, for $x \geq 0,$ and reflecting it across the $y$ axis.
In the following, therefore, we will assume that $x \geq 0.$
Next, Equation $(1)$ implies that $$y^2 = \frac{b^2}{a^2} x^2 - b^2. $$ Also, $c = \sqrt{a^2 + b^2}$ implies that $a^2 + b^2 = c^2.$ Therefore \begin{align} (x\pm c)^2 + y^2 &= x^2 \pm 2cx + c^2 + \frac{b^2}{a^2} x^2 - b^2 \\ &= \left(1 + \frac{b^2}{a^2}\right) x^2 \pm 2cx + c^2 - b^2 \\ &= \frac{a^2 + b^2}{a^2} x^2 \pm 2cx + a^2 \\ &= \frac{c^2}{a^2} x^2 \pm 2cx + a^2 \\ &= \left(\frac ca x \pm a\right)^2. \\ \end{align}
Now, $c = \sqrt{a^2 + b^2}$ also implies $c \geq a,$ and Equation $(1)$ implies that $$ \frac{x^2}{a^2} = 1 + \frac{y^2}{b^2} \geq 1, $$ which implies that $x^2 > a^2,$ which implies $x > a$ (since we have assumed that $a > 0$ and $x \geq 0$). Therefore $\frac ca x \geq x \geq a$. It follows that $$\sqrt{\left(\frac ca x - a\right)^2} = \left\lvert\frac ca x - a\right\rvert = \frac ca x - a,$$ and of course $$\sqrt{\left(\frac ca x + a\right)^2} = \left\lvert\frac ca x + a\right\rvert = \frac ca x + a.$$
Now we put these facts together. The distance from $(x,y)$ to one focus is $\sqrt{(x+c)^2+y^2}$ and the distance to the other focus is $\sqrt{(x-c)^2+y^2},$ so the absolute difference of these distances is
\begin{align} \left\lvert\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}\right\rvert &= \left\lvert\sqrt{\left( \frac ca x + a\right)^2} - \sqrt{\left( \frac ca x - a\right)^2}\right\rvert \\ &= \left\lvert\left(\frac ca x + a\right) - \left(\frac ca x - a\right)\right\rvert\\ &= 2a, \end{align}
showing that $(x,y)$ is on the hypberbola with foci at $(c,0)$ and $(-c,0)$ such that the absolute difference of distances to the foci is $2a.$
In conclusion, Equation $(1)$ is satisfied by the points on that hyperbola and only by the points on that hyperbola.