The equation of the normal to the graph of the function y=y(x) at a point with abscissa x0 is the following:
y−y(x0)=−y′(x0)1(x−x0)
In our case, y=xc2, x0=ct, y(x0)=tc. It follows that y′=−x2c2. Therefore, the equation of the normal at point x0=ct is
y−tc=−−c2t2c21(x−ct) which is equivalent to
y−tc=t2(x−ct) and to
ty−c=t3x−ct4.
Finally, we have the equation of the normal in the following form:
t3x−ty=c(t4−1)
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