Question #282731

a2x2=y3(2a-y) trace the curve


1
Expert's answer
2021-12-28T16:45:09-0500

We shall show the trace of the equation

a2x2=y3(2ay)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 x2=0x^{2}=0     \implies x=0,0x=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).(0,2a). From the equation of the curve, we have

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

For y<0y<0 or y>2ay>2a, xx is imaginary.

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







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