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)y = y(x) at a point with abscissa x0x_0 is the following:


yy(x0)=1y(x0)(xx0)y-y(x_0)=-\frac{1}{y'(x_0)}(x-x_0)


In our case, y=c2x,  x0=ct, y(x0)=cty=\frac{c^2}{x}, \ \ x_0=ct,\ y(x_0)=\frac{c}{t}. It follows that y=c2x2y'=-\frac{c^2}{x^2}. Therefore, the equation of the normal at point x0=ctx_0=ct is


yct=1c2c2t2(xct)y-\frac{c}{t}=-\frac{1}{-\frac{c^2}{c^2t^2}}(x-ct) which is equivalent to


yct=t2(xct)y-\frac{c}{t}=t^2(x-ct) and to


tyc=t3xct4ty-c=t^3x-ct^4.


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


t3xty=c(t41)t^3x-ty=c(t^4-1)


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