Find the curve whose tangent lines in every point intersect $x$ and $y$ axis with property that sum of intersection lengths on $x$ and $y$ axis are equal to a constant $a$.
I'll show you my way of solving this problem and I'd be grateful if someone can confirm it's the right way to solve this problem.
Let $M(x,y)$ be the point on the curve where our tangent line crosses and let $L, N$ be two points where tangent line intersects $y$ and $x$ axis respectively.
Let's find coordinates of $L$ and $N$. Write the tangent line equation: $Y – y(x) = y'(x) ( X – x)$
For $Y = 0$ we can find coordinates of N, putting $Y=0$ in the equation above we get $N(x-\frac{y}{y'},0)$.
Same for $L$ point, if we put $X=0$ we get $L(0,-xy'+y)$.
Now we have $N(x-\frac{y}{y'},0),M(x,y),L(0,-xy+y')$. Let's use the length property.
$|LO|+|NO|=a$, where $O$ is center of coordinate system $(0,0)$. I get the following equation:
$y^2-2xyy'+x^2(y')^2+x^2-2x\frac{y}{y'}+\frac{y^2}{(y')^2}=0$ and if I solve this I'll get the equation of curve.
Can you confirm that this is the right way to solve this problem or are there any easier ways to do this?
Best Answer
The ODE you obtained looks pretty awful. Its solution is the family $(\ell_c)_{c\in{\mathbb R}}$ of lines you worked with, plus a certain "extra" solution $\epsilon$ that can only be obtained with an extra procedure, even if you could solve this ODE. There is an easier way:
When you draw a few lines $\ell_c$ having axes intercepts summing to the given $a$ you will see that these lines are tangents to a common curve $\epsilon$. This curve, called envelope of the family of lines, is your animal. (It so happens that exactly your example is treated in the linked Wikipedia entry.)
To solve the problem you have to write up the equation of your line family. It is given by $$\ell_c:\qquad{x\over c}+{y\over a-c}=1\quad(0<c<a)\ ,$$ or $$(a-c)x+c y-c(a-c)=0\ .\tag{1}$$ There is an "obscure" general procedure to obtain the envelope of a given family of curves. You have to differentiate $(1)$ partially with respect to the parameter $c$, and obtain a second equation $(2)$. Solve the system $(1)\wedge(2)$ for $x$ and $y$ as functions of $c$, and you have a parametrization of the envelope $\epsilon$. This proceeding of course has theoretic reasons which I don't explain here.
But your example is so simple that we can obtain the envelope without calculus: The family $(\ell_c)_{c\in{\mathbb R}}$ does not cover the full $(x,y)$-plane. There are points $(x,y)$ that are passed by two lines $\ell_{c_1}$, $\ell_{c_2}$, and there are points $(x,y)$ that are not passed by a line $\ell_c$. The boundary between the two parts is exactly the envelope $\epsilon$. We can therefore do the following: Solve the equation $(1)$ for $c$. This is a quadratic equation with discriminant $$D(x,y)=(y-x-a)^2-4ax\ .$$ When $D(x,y)>0$ we obtain two real $c$-values $c_1$, $c_2$. When $D(x,y)<0$ there is no real solution $c$. The points of $\epsilon$ are exactly the points satisfying $D(x,y)=0$, so that we can write $$\epsilon:\quad(y-x-a)^2-4ax=0\ .$$ This is a parabola opening up in direction north-east.