The gradient of the normal to a curve at any point is equal to the $x$- coordinate at that point. If the curve passes through the point $(e^2,3)$, find the equation of the curve in the form $y=\ln (g(x))$, where $g(x)$ is a rational function, $x>0$.
From the information given the gradient of the normal is $e^2$ it follows that the gradient of the tangent is $\frac{-1}{e^2}$. I can also find the equations of the tangent by using the point given $y=\frac{-x}{e^2}$+4. I am stuck after this. I don't know how to find the equation of the gradient. After that I should integrate this equation and I will get the result I am looking for. If I am incorrect then please tell me.
Best Answer
The slope of the normal line is $-1/y'(x)$. Since this is supposed to be equal to $x$, we have $y'(x)=-1/x$. Integration yields $y(x)=-\ln x+C$. From the given data $C=5$. Hence $y= \ln (e^5 /x)$.