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

Question #157491
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)
1
Expert's answer
2021-02-02T05:21:25-0500

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)"


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