$C$ is a curve of $y(x)$ function on $XY$ , there is point $P:=(x,y)$. A parallel line and perpendicular line of C pass through the point $P$.

$AB=Y = y + y'(x) (X-x)$

$PG= Y = y - \frac{1}{y'(x)} (X-x)$

Find the curves such that $PM=MG$. I need to find the formula of the curves family and don't know how to do it. enter image description here

My solution :

I will find $M$ and $G$:

$0=y - \frac{1}{y'(x)} (X-x) \implies X_G=yy'(x)+x \implies G=(yy'(x)+x,0).$

$Y_M = y - \frac{1}{y'(x)} (0-x) \implies Y_M=y+\frac{x}{y'(x)} \implies M=(0,y+\frac{x}{y'(x)}).$

Find $PM : \sqrt{(x-0)^2+(y-(y+\frac{x}{y'(x)})^2} = \sqrt{x^2+(\frac{x}{y'(x)})^2}$

Find $MG$ : $\sqrt{(0-(yy'(x)+x))^2+(y+\frac{x}{y'(x)}-0)^2}$

$PM=MG : \sqrt{x^2+(\frac{x}{y'(x)})^2}= \sqrt{(0-(yy'(x)+x))^2+(y+\frac{x}{y'(x)}-0)^2}$

$ \implies x^2+(\frac{x}{y'(x)})^2=(y\cdot y'(x)+x)^2+(y+\frac{x}{y'(x))}^2$

$ \implies x^2 + \frac{x^2}{(y'(x))^2}=y^2\cdot (y'(x))^2+2y\cdot y'(x)\cdot x+x^2+y^2+\frac{2xy}{y'(x)}+\frac{x^2}{(y'(x))^2} $

$\implies 0=y^2\cdot y'(x)+2y\cdot y'(x)\cdot x+y^2+\frac{2xy}{y'(x)} \implies y'(x)=\sqrt{\frac{x}{-y-x}}.$

Is this part of my solution correct?

  • $\begingroup$ I react to your last equation :$y'(x)=\sqrt{\frac{x}{-(y+x)}}$ is impossible because under the root sign you have a negative quantity if your point is the first quadrant $x>0, y>0$. $\endgroup$
    – Jean Marie
    Nov 20, 2021 at 11:35

2 Answers 2


Here is an approach that follows the line of reasoning you demonstrate in your post.

The equation of line normal to your unknown curve $C$ as the point $P=(a,f(a))$ is $$y=-\frac{1}{f'(a)}(x-a)+f(a)$$ By finding the $x$ and $y$ intercepts of this line we determine the points $M$ and $G$ to be $$M=\Big(0,f(a)+\frac{a}{f'(a)}\Big)$$ $$G=\Big(f'(a)f(a)+a,0\Big)$$ Then $$(PM)^2=(GM)^2 \iff f'(a)=-{2a \over f(a)}$$ In other words, your unknown function $f$ must satisfy the differential equation $y'=-\frac{2x}{y}$ providing you with the same solution curves as @JeanMarie.

  • $\begingroup$ It's very well to show, as you have done, that one can follow the track already begun. I recognize that my solution was influenced by the fact that I hald already solved a few subtangent and subnormal problems... $\endgroup$
    – Jean Marie
    Nov 22, 2021 at 0:12

It is difficult for me to find another error than the one I have already found.

In fact, there is a simpler approach.

Let me say first what is the solution, the proof being given hereafter.

The solution is any ellipse (at least its arc in the first quadrant) with implicit equation :

$$x^2+\frac{y^2}{2}=k \tag{1}$$

or, using a parametric representation:

$$\begin{cases}x&=&\cos(t)\\y&=&\sqrt{2} \sin(t)\end{cases}$$

Here is a Desmos representation where $k=1$ that one can animate by playing on the cursor for $s$.

enter image description here

Now, let us give the proof.

Let $P'$ be the projection of $P$ onto the $x$ axis. The length of line segment $GP'$ is classically called the "subnormal" of the curve ; it is given by a classical formula $|yy'|$. Taking into account the fact that $(x,y)$ is in the first quadrant $x>0, \ y>0$ and that the slope is negative, we have: $|yy'|=-yy'.$

The important remark now is that we need only consider abscissas. Indeed, condition $PM=MG$ gives by projection $P′O=OG$ ($O$ being the origin) itself equivalent to $$GP'=2OP' \ \ \ \iff \ \ \ -yy'=2x,$$

a differential equation whose integration gives (1), $k$ being an integration constant.

Remark: the solution curve, defined at first for $x>0,y>0$ is in fact valid for $|x|>0,|y|>0$.

Historical corner: issues on subnormals and subtangents were rather common in the 17th and 18th centuries. They were sometimes used as challenges between mathematicians just before and still after the beginning of Calculus.

  • $\begingroup$ Thank you ! but how we know $PM=MG$ exists in the solution? $\endgroup$
    – algo
    Nov 20, 2021 at 13:55

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