Answer to Question #282731 in Calculus for fahad

Question

a²x²=y³(2a-y) trace the curve

Answer 1

We shall show the trace of the equation

$a^{2}x^{2}=y^{3}(2a-y)$

(1) The curve is symmetrical about y-axis

(2)It passes through the origin and the tangents at the origin are $x^{2}=0$ $\implies$ $x=0,0$

(3) The curve has no asymptote

(4) The curve meets the y-axis at the origin only and meets the y-axis at $(0,2a).$ From the equation of the curve, we have

$$x=\frac{y}{a}\sqrt{y(2a-y)}$$

For $y<0$ or $y>2a$, $x$ is imaginary.

Thus, the curve entirely lies between $y=0$ $x-axis$ and $y=2a,$ which is shown in the below figure