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Find a curve such that the surface of the triangle bounded by the line going through the tangent point and perpendicular to the x-axis and the tangent line to the graph is equal to $a^2$.

I didn't understand the question at first glance, and I found related answers like this one - but this question also assumes the triangle is bounded by a perpendicular line to the $x$ axis.

This is what I got so far, but I don't think I got the DE entirely correct. I think I need some help with interpreting the question.


The tangent line to the curve at any point $x$ is given by:

$y-xy^{\prime}=0$

This line intersects the $x$ axis at some point $x_{0}$ s.t. $y(x_0)=0$. Then, for any $x>x_{0}$, a $\perp$ line of height $y\left(x\right)$ intersects the tangent line.

So the triangle is defined by:

$A\left(x_{0},0\right)$
$B\left(x,0\right)$
$C\left(x,y-xy^{\prime}\right)$

Thus, the area of $\triangle ABC$ is given by:

$\frac{1}{2}\left(x-x_{0}\right)\left(y-xy^{\prime}\right) =a^{2}$

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    $\begingroup$ Are you sure the equation of tangent is correct? According to me, it should be $\frac{y-Y}{x-X}=g'(X)$ where $(X,Y)$ is the point of tangency and $g(x)$ is the desired curve. $\endgroup$ – Shubham Johri Nov 1 '20 at 20:49
  • $\begingroup$ @ShubhamJohri But isn't the point of tangency should be a general point, not a particular one? How does $(X,Y)$ differ from the $(x,y)$ arguments in your equation? $\endgroup$ – gbi1977 Nov 1 '20 at 20:58
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Let the point of tangency on the curve $y=g(x)$ be $(X,g(X))$.

Equation of tangent is $\frac{y-g(X)}{x-X}=g'(X)$.

$x-$intercept of tangent is $X-g(X)/g'(X)$.

The area of the triangle in question is $\frac12\times|X-(X-g(X)/g'(X))|\times|g(X)|=a^2$.

This gives the ODE$$|g'|=g^2/2a^2$$

Since we want to find any one solution, we let $g'\ge0$. Thus$$\int\frac{dg}{g^2}=-\frac1g=\frac x{2a^2}+c$$You can assume $g'\le 0$ and you would get $1/g=x/2a^2+c_1$ which is also admissible.

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  • $\begingroup$ I don't follow, why x−intercept of tangent is $X−\frac{g(X)}{g′(X)}$? $\endgroup$ – gbi1977 Nov 1 '20 at 21:07
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    $\begingroup$ Just substitute $y=0$ in tangent eqn. @gbi1977 $\endgroup$ – Shubham Johri Nov 1 '20 at 21:08
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    $\begingroup$ @gbi1977 Just remember that at $x=-2a^2c$ the function is not defined. So remove that from the domain. $\endgroup$ – Shubham Johri Nov 1 '20 at 21:11

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