\(y' = \frac{3y}{x}\) is normal
\(\because\) Condition of normal is y1y2 = -1
Let m is slope of curve whose normal is \(y' = \frac{3y}{x}\).
\(\therefore \) my1 = -1
⇒ \(m = \frac{1}{y^1} = \cfrac{-1}{\frac{3y}{x}} = \frac{-x}{3y}\)
⇒ \(\frac{dy}{dx} = \frac{-x}{3y}\) \((\because m = \frac{dy}{dx})\)
⇒ 3y dy = -x dx
⇒ \(\frac{3y^2}{2} = \frac{-x^2}{2} + \frac c2\) , where \(\frac c2\) is integral constnat
⇒ 3y2 + x2 = c
Interval i = {(x, y) : 3y2 +x2 = c}.
