Answer to Question #153127 in Calculus for Maneeka

Question

Trace the curve x3 + y2 = 2axy 

Answer 1

Equation of the curve :- $\pmb {x^3+y^2=2axy}$

$x^3+y^2=2axy\\ \Rightarrow y^2-2axy+x^3=0\\ \Rightarrow y=\frac{2ax \pm \sqrt{4a^2x^2-4x^3}}{2}=ax\pm x\sqrt{a^2-x}$

$\therefore$ The curve is combination of two Branches :

$y=ax+ x\sqrt{a^2-x}$ ...[Branch 1]

and, $y=ax- x\sqrt{a^2-x}$ ...[ Branch 2]

Differentiating w.r.t x,

$\frac{dy}{dx}=a\pm \left(\sqrt{a^2-x} -\frac{x}{2\sqrt{a^2-x} }\right)$

Symmetry :-

The curve is not symmetric w.r.t. any line or axis.

Region :-

$y=ax \pm x\sqrt{a^2-x} \\ y \in \R \Rightarrow a^2-x >0 \Rightarrow x<a^2$

$\therefore x \in (-\infty,a^2), y\in (-\infty, \infty)$

Point of Intersection :-

Putting $x=0,$ $y=0$

Putting $y=0,$ $x=0$

$\therefore$ The curve passes through origin.

Let $(h,k)$ be the intersection of Branch 1 and Branch 2.

$\therefore ah+h\sqrt{a^2-h}=ah-h\sqrt{a^2-h}\\ \Rightarrow h\sqrt{a^2-h}=0\\ \Rightarrow h= 0 \; or, \; a^2\\ \Rightarrow k=0 \; or, \; a^3$

So, Branch 1 and Branch 2 meet at $(0,0)$ and $(a^2,a^3)$.

Tangents at Origin :-

At $(0,0) , \frac{dy}{dx}=a\pm \left(\sqrt{a^2-0}-0\right)=a\pm a =2a\; or, \; 0$

$\therefore$ Tangents at origin are :-

$y=2ax \; \;\;and\;\;\; y=0$

So, $(0,0) \; is \; a \; node.$ [ As two tangents are distinct ]

Asymptotes :-

Coefficient of highest degree of $x$ and $y$ are both constant. So, There does not exist any horizontal or vertical asymptote.

Putting $y=mx+c,\\$

$x^3+m^2x^2+2mcx+c^2=2axy$

$\Rightarrow$ Coff. of $x^3=1=Constant$

So There does not exist any oblique asymptote either.

Derivatives :-

$y=ax\pm x\sqrt{a^2-x} \\ \Rightarrow \frac{dy}{dx}=a\pm \left(\sqrt{a^2-x} -\frac{x}{2\sqrt{a^2-x} }\right)\\$

Now, $\frac{dy}{dx}=0$

$\Rightarrow a= \pm \frac{2(a^2-x)-x}{2\sqrt{a^2-x} }\\ \Rightarrow4a^2(a^2-x)=4a^4-12a^2x+9x^2\\ \Rightarrow x=0 \; or, \; x=\frac{8a^2}{9}$

$\therefore \frac{dy}{dx}=0$ at $x=8a^2/9$ for Branch 1

at $x=0$ for Branch 2

Now,

for $\pmb{a>0, }$

$$Branch \;1$$

$y=ax+ x\sqrt{a^2-x} \\\Rightarrow \frac{d^2y}{dx^2}= \frac{3x-4a^2}{4(a^2-x)^{3/2}}$

At $x=\frac{8a^2}{9}, \; \frac{d^2y}{dx^2}=-\frac{9}{a}<0$

So, Branch 1 has a local maxima at $x=\frac{8a^2}{9}$ .

$$Branch \;2$$

$y=ax- x\sqrt{a^2-x} \\\Rightarrow \frac{d^2y}{dx^2}= -\frac{3x-4a^2}{4(a^2-x)^{3/2}}$

At $x=0, \; \frac{d^2y}{dx^2}=\frac{1}{a}>0$

So, Branch 2 has a local minima at $x=0$ .

$$\pmb{(a>0)}$$

for $\pmb{a<0,}$

$$Branch \;1$$

$y=ax+ x\sqrt{a^2-x} \\\Rightarrow \frac{d^2y}{dx^2}= \frac{3x-4a^2}{4(a^2-x)^{3/2}}$

At $x=\frac{8a^2}{9}, \; \frac{d^2y}{dx^2}=-\frac{9}{a}>0$

So, Branch 1 has a local minima at $x=\frac{8a^2}{9}$ .

$$Branch \;2$$

$y=ax- x\sqrt{a^2-x} \\\Rightarrow \frac{d^2y}{dx^2}= -\frac{3x-4a^2}{4(a^2-x)^{3/2}}$

At $x=0, \; \frac{d^2y}{dx^2}=\frac{1}{a}<0$

So, Branch 2 has a local maxima at $x=0$ .

$$\pmb{(a<0)}$$