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 \[Y - y = - \frac{1}{\frac{dy}{dx}}\left( X - x \right)\]

It is given that the curve passes through the point (3, 0). Then,
\[0 - y = - \frac{1}{\frac{dy}{dx}}\left( 3 - x \right)\]
\[ \Rightarrow - y = - \frac{1}{\frac{dy}{dx}}\left( 3 - x \right)\]
\[ \Rightarrow y\frac{dy}{dx} = 3 - x\]
\[ \Rightarrow y dy = \left( 3 - x \right)dx\]
\[ \Rightarrow \frac{y^2}{2} = 3x - \frac{x^2}{2} + C . . . . . \left( 1 \right)\]
\[\text{ Since the curve passes through the point }\left( 3, 4 \right), \text{ it satisfies the equation .} \]
\[ \Rightarrow \frac{4^2}{2} = 3\left( 3 \right) - \frac{3^2}{2} + C\]
\[ \Rightarrow C = 8 - 9 + \frac{9}{2}\]
\[ \Rightarrow C = \frac{9}{2} - 1 = \frac{7}{2}\]
\[\text{ Putting the value of C in }\left( 1 \right),\text{ we get }\]
\[\frac{y^2}{2} = 3x - \frac{x^2}{2} + \frac{7}{2}\]
\[ \Rightarrow y^2 = 6x - x^2 + 7\]
\[ \Rightarrow x^2 + y^2 - 6x - 7 = 0\]