Find the equation of the curve so drawn that every point on it is equidistant from the origin and the intersection of the x-axis with the normal to the curve at the point.
Solution:
Plot points on the curve,
A(x_{1},y_{1})We all know that a Slope of a Tangent corresponds to m, and its negative reciprocal is equal to the Slope of a Normal. Thus, we use Point-Slope Formula.
y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right)As the normal intersects the x-axis, y = 0
Substituting to the previous equation, we get
\begin{align*}
-y_{1}&=-\frac{1}{m}\left(x-x_{1}\right)\\
-my_{1}&=-1\left(x-x_{1}\right)\\
-my_{1}&=-x+x_{1}\\
x&=x_{1}+my_{1}
\end{align*}
By using distance formula, from the origin (0,0), to point (x1+y1) = from intersection to (x1+y1)
\begin{align*}
\sqrt{\left(x_{1}-0\right)^2+\left(y_{1}-0\right)^2}&=\sqrt{\left(x_{1}+my_{1}-x_{1}\right)^2+\left(0-y_{1}\right)^2}\\
\sqrt{x_{1}^2+y_{1}^2}&=\sqrt{m^2y_{1}^2+y_{1}^2}\\
x_{1}^2+y_{1}^2&=m^2y_{1}^2+y_{1}^2\\
x_{1}^2&=m^2y_{1}^2\\
x_{1}&=m_{1}y_{1}\:\:\:\:;m=\frac{dy}{dx}\\
x_{1}&=\frac{dy}{dx}y_{1}
\end{align*}
Change x1 and y1 to x and y,
\begin{align*}
x&=y\frac{dy}{dx}\\
xdx&=ydy
\end{align*}
By integrating,
\begin{align*}
\int \:ydx&=\int \:xdy\\
\frac{y^2}{2}&=\frac{x^2}{2}+C\\
y^2&={x^2}+2C\\
\end{align*}
We get,
y^2-x^2-=2C
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