Answer to Question #157491 in Differential Geometry | Topology for Vanessa

Question

Prove that the equation normal to the rectangular hyperbola xy = c^2 at the point P(ct, c/t) is t^3*x - ty = c(t^4 - 1)

Answer 1

The equation of the normal to the graph of the function $y = y(x)$ at a point with abscissa $x_0$ is the following:

$y-y(x_0)=-\frac{1}{y'(x_0)}(x-x_0)$

In our case, $y=\frac{c^2}{x}, \ \ x_0=ct,\ y(x_0)=\frac{c}{t}$. It follows that $y'=-\frac{c^2}{x^2}$. Therefore, the equation of the normal at point $x_0=ct$ is

$y-\frac{c}{t}=-\frac{1}{-\frac{c^2}{c^2t^2}}(x-ct)$ which is equivalent to

$y-\frac{c}{t}=t^2(x-ct)$ and to

$ty-c=t^3x-ct^4$.

Finally, we have the equation of the normal in the following form:

$t^3x-ty=c(t^4-1)$