Find the curves for which the area between the x -axis, the…
Find the curves for which the area between the $x$-axis, the tangent line, and the radius vector to the point of tangency equals $a^2$.
Explanation of Method
Express the triangle's area in terms of $x,y,y'$, set it equal to $a^2$, and solve the resulting linear equation for $x$ as a function of $y$.
Step 1: Set up the area condition
The tangent at $(x,y)$ meets the $x$-axis at $X=x-y/y'$. The triangle $OAP$ (origin, $x$-intercept, point of tangency) has area $\dfrac12\left(xy-\dfrac{y^2}{y'}\right)$. Setting this equal to $a^2$:
\[xy-\frac{y^2}{y'}=2a^2.\]
Step 2: Rewrite as linear in $x(y)$
Solving for $\dfrac{dx}{dy}=1/y'$: $\dfrac{dx}{dy}-\dfrac{x}{y}=-\dfrac{2a^2}{y^2}$.
Step 3: Solve the linear equation
With integrating factor $\mu(y)=1/y$: $\dfrac{d}{dy}\!\left(\dfrac{x}{y}\right)=-\dfrac{2a^2}{y^3}$, so $\dfrac{x}{y}=\dfrac{a^2}{y^2}+C$, i.e. $x=\dfrac{a^2}{y}+Cy$.
Result
Final answer:
\[\boxed{\,xy=a^2+Cy^2,\qquad C\in\mathbb{R}}\]