Orthogonal and isogonal trajectories of tangent for hyperbola and ellipse.

Question

I'm trying to understand a task from DIFFERENTIAL and INTEGRAL CALCULUS 102. Find curves for which the product of the distance of any tangent line to two given points is constant. Answer: Ellipses and hyperbolas. (Orthogonal and isogonal trajectories.) What did Piskunov mean under two given points? Any thoughts will be appreciated.

Answer 1

Further thoughts based on my comment above

Let the two given points be $(\pm c,0)$, and the tangent be $y=mx+k$,

Now

$$\frac{mc+k}{\sqrt{1+m^2}} \times \frac{-mc+k}{\sqrt{1+m^2}}= \pm b^2$$

Positive (constant product) when the points are on the same side of the tangent whereas negative are opposite.

$$\frac{k^2-m^2c^2}{1+m^2}=\pm b^2$$

$$k^2=m^2(c^2 \pm b^2) \pm b^2$$

Take $a^2=c^2 \pm b^2$,

$$(y-mx)^2=a^2m^2 \pm b^2$$

which is known as the magical equation to the tangent of an ellipse $(+)$ or a hyperbola $(-)$ with slope $m$. The curve is an envelope of a family of tangents, namely

$$F(x,y)\equiv (mx-y)^2-(a^2 m^2 \pm b^2)=0$$

Solving $$\frac{\partial F}{\partial m}=F=0$$

will give the curves.

Answer 2

After @Ng Chung Tak

$mx-y=\pm \sqrt{a^2m^2\pm b^2}$. Next, let us take only upper signs, then $y=mx-\sqrt{a^2m^2+b^2}, m=dy/dx$, this is a Clairaut equation $$y=xy'-\sqrt{a^2 y'^2+b^2}~~~~(1)$$ representing the family of tangents to a fixed curve ; the fixed curve is obtained by differentiating (1) w.r.t. $x$, we get $$x-\frac{a^2y'}{\sqrt{a^2y'^2+b^2}}=0~~~(2)$$ apart from the general equation $y''=0$. Solving (2), we get $$\frac{dy}{dx}=\frac{bx}{a\sqrt{a^2-x^2}} \implies y=\frac{b}{a}\int \frac{x dx}{\sqrt{a^2-x^2}} \implies y=\frac{b}{a} \sqrt{a^2-x^2}.$$ $$\implies \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$ which is an ellipse. Similarly, by considering lower signs above for $\pm$, we can get a hyperbola.

For Clairaut equation see: https://en.wikipedia.org/wiki/Clairaut%27s_equation