Answer to Question #146275 in Calculus for Sourav Mondal

Question

Trace the curve (x²+y²)x=ay², a>0 stating all the properties used in the process

Answer 1

$\displaystyle \textbf{\textsf{Intercepts}}\\ (x^2+y^2)x=ay^2\\ \textsf{At}\,\, x = 0, \,y = 0\\ \therefore\textsf{The curve passses through}\\ \textsf{the origin.}\\ \textbf{\textsf{Stationary points}}\\ x^3 + y^2x = ay^2\\ 3x^2 + y^2 + 2xyy' = 2ayy'\\ \frac{3x^2 + y^2}{2ay - 2xy} = y'\\ \textsf{The value of}\,x \, \textsf{for which the}\\ \textsf{there is is a stationary point}\, y' = 0\\ 3x^2 = -y^2\\ y = j\sqrt{3}x\, \textsf{where}\, j = \sqrt{-1}\\ \textsf{For the line}\, y = j\sqrt{3}x\,, \textsf{for all values}\\ \textsf{of}\,x, \textsf{there are no}\\ \textsf{values of}\, y \, \textsf{that exists}\\ \textsf{in the real plane and vice-versa}\\ \textsf{except at}\, (x, y) = (0, 0). \\ \textsf{Therefore, there is no maximum value or minimum value}\\ \textsf{of}\,y\, \textsf{and there is a stationary point}\, x = 0. \, \\ \textsf{Since at}\, \frac{\mathrm{d}y}{\mathrm{d}x} = 0\,\, \textsf{at}\,\, x = 0.\\ \textbf{\textsf{Asymptotes}}\\ x^3 + y^2x = ay^2\\ x^3 = y^2(a - x)\\ y = \pm\sqrt{\frac{x^3}{a - x}}\\ \textsf{The line}\,\,x = a \, \, \textsf{is a vertical asymptote}\\ \textsf{And as}\,\, x \rightarrow a, y \rightarrow \pm\infty.\\ \textsf{Therefore, the graph is increasing}\\ \textsf{and decreasing for}\,\, x > 0.$