Finding a curve - first order ODE 
Find a curve whose tangent lines create a triangle of area $2a^2$ with the $x-$ and $y$-axes.

First off, the tangent line equation is:$$y=f'(X)(x-X)+Y$$ where $(X,Y)$ is an arbitrary point. I realize that now I have to evaluate it at $X=0$ and $Y=0$, and then multiply it and the absolute value of that must equal $4a^2$.
But I dont know how to translate what I wrote into an equation.
Im sorry for my English, its not my native language, and I try my best translating my tasks and what Im trying to do. Feel free to correct what I wrote if there are any linguistic mistakes.
 A: The concept you need is the envelope of a family of curves. 
Given a family of curves $C_t$ parametrized by a parameter $t$. One definition of the envolope is a curve tangent
to all of the $C_t$. If you view the tangent lines of the curve you want as a family of curves, the curve you seek is the corresponding envelope.
Let's say the family of curves is given implicitly by some function $F(t,x,y)$,
$$(x,y) \in C_t\quad\iff\quad F(t,x,y) = 0$$
The envelope for $\{\;C_t\;\}$ will be given by the set of points for which
$$F(t,x,y) = \frac{\partial}{\partial t}F(t,x,y) = 0$$
Let's use the curve you seek as an example. If you are given a line of the from
$$\frac{x}{x_0} + \frac{y}{y_0} = 1\quad\text{ with }\quad x_0, y_0 > 0$$
The area it forms with the $x$ and $y$ axes are given by $\frac12 x_0 y_0$. 
If this is equal to $2a^2$, then we can find some $t$ such that 
$\displaystyle\;x_0 = \frac{2a}{t}$ and $y_0 = 2a t$. 
We can parametrize the set of tangent lines as
$$C_t :  F(t,x,y) = xt  + \frac{y}{t} - 2 a = 0$$
The condition $\displaystyle\;\frac{\partial}{\partial t}F(t,x,y) = 0\;$ becomes
$$x - \frac{y}{t^2} = 0 \quad\implies\quad t = \sqrt{y/x}$$
Substitute this back into the condition $\displaystyle\;F(t,x,y) = 0\;$, one get
$$x\sqrt{y/x} + y \sqrt{x/y} - 2a = 0 \quad\iff\quad xy = a^2$$
The curve you seek is that portion of the hyperbola $xy = a^2$ in the first quadrant.
A: it may help to appreciate Achille's elegant answer if i append a more elementary treatment starting from Lugi's equation for the tangent. if $(X,Y)$ is a generic point on the solution curve, we will denote the derivative of $Y$ wrt $X$ by $Y'$ the tangent at $(X,Y)$ is then:
$$
y-Y=Y'(x-X)
$$
thus the points of intersection with the axis, $(x_0,0)$ and $(0,y_0)$ are given by:
$$
y_0= Y - XY' \\
x_0= - \frac{y_0}{Y'}
$$
and the area-of-triangle condition is:
$$
\frac12 x_0y_0 = 2a^2
$$
substituting we have
$$
(Y-XY')^2 = -4a^2Y'
$$
and upon differentiating w.r.t $X$ this becomes
$$
2(Y-XY')(Y'-Y'-XY'') = -4a^2Y''
$$
i.e.
$$
(Y-XY')XY'' = 2a^2Y''
$$
in a region where $Y'' \ne 0$ this may be written:
$$
- \left(\frac{Y}{X}\right)' = 2a^2 X^{-3}
$$
which integrates to give
$$
\frac{Y}{X} = a^2X^{-2} + c
$$
if $X \to \infty$ the triangle condition determines that $Y \to 0$ hence $c=0$ giving
$$
XY = a^2
$$
A: We find singular solution of single parameter straight line family of Clairaut's type. The powerful C discriminant (and also P  discriminant ) are so useful, as  also mentioned by achilles  hui. 
The required singular solution or envelope is the eliminant between F(x,y,u) = 0 and its partial derivative with respect to parameter u.
$ x/u + y/v = 1, u v = 4 a^2 $
Plug in v of second into first equation to eliminate it, $ x/u + y *u/ 4 a^2 = 1 ..... (1) $ 
Partial differentiation w.r.t. u gives$ - x/ u^2 + y/ 4 a^2 = 0 ..... (2)$
Eliminate u from (1) and (2) and simplify to get $ x y = a^2$, the desired curve as a rectangular hyperbola touching all straight lines with constant intercept product.
You can play with intercepts. For example if you want $ u^2 + v^2 = L^2 = constant$ where a ladder length L slides down y-axis keeping bottom end along x-axis, following the same steps above you get an astroid 
$ x^ {2/3} + y^ {2/3} = L^ {2/3 } $ .
