Question

The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.

Answer 1

Let P (x, y) be any point on the curve. The equation of the normal at P (x, y) to the given curve is given as
$Y-y=-\frac{1}{{\displaystyle \frac{dy}{dx}}}\left(X-x\right)$
It is given that the curve passes through the point (3, 0). Then,
$$\begin{gathered} 0-y=-\frac{1}{{\displaystyle \frac{dy}{dx}}}\left(3-x\right) \\ \Rightarrow -y=-\frac{1}{{\displaystyle \frac{dy}{dx}}}\left(3-x\right) \\ \Rightarrow y\frac{dy}{dx}=3-x \\ \Rightarrow ydy=\left(3-x\right)dx \\ \Rightarrow \frac{y^2}{2}=3x-\frac{x^2}{2}+C.....\left(1\right) \\ \mathrm{Since}\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\left(3,4\right),\mathrm{it}\mathrm{satisfies}\mathrm{the}\mathrm{equation}. \\ \Rightarrow \frac{4^2}{2}=3\left(3\right)-\frac{3^2}{2}+C \\ \Rightarrow C=8-9+\frac{9}{2} \\ \Rightarrow C=\frac{9}{2}-1=\frac{7}{2} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \frac{y^2}{2}=3x-\frac{x^2}{2}+\frac{7}{2} \\ \Rightarrow y^2=6x-x^2+7 \\ \Rightarrow x^2+y^2-6x-7=0 \end{gathered}$$