Answer to Question #108857 in Calculus for Cifal Shaul

Question

Find the area included between the curve x^3 + y^3 = 3axy and its asymptote

Answer 1

$x^3+y^3=3axy$

$[1+(\frac{y}{x})^3]=\frac{3a}{x}(\frac{y}{x})$

$\frac{[1+(\frac{y}{x})^3]}{(\frac{y}{x})}=\frac{3a}{x}$

As x tends to $\infty$ and $-\infty$ ; RHS tends to 0,so LHS also tend to 0.

So , $1+(\frac{y}{x})^3$ tends to 0,thus,$\frac{y}{x}=-1$

Hence the asymptote has slope −1 and so is of the form $y+x=k$

Substituting $y=k-x$ in the equation of curve.

$−3k²x+3kx²−k³ = 3ax(k-x)$

$−3k²/x+3k−k³/x² = 3a[(k/x)-1 ]$

$k=-a$

This,equation of asymptote is $x+y+a=0$

The product of slopes of the line symmetrical to $y=x$ and the equation of asymptote $x+y+a=0$ is (-1) i.e. they are at right angle to each other.Whence if the axes of references are turned through an angle of 45° ,the new X axis coincides with the symmetrical line $y=x$ with a changed equation of curve having asymptote parallel to the new Y-axis instead the oblique asymptote $x+y+a=0$ .

After transformation new X and Y are:

$X=xcos(45°)-ysin(45°)=\frac{x-y}{\sqrt{2}}$

$Y=ysin(45°)+ycos(45°)=\frac{x+y}{\sqrt{2}}$

$x=\frac{X+Y}{\sqrt{2}};y=\frac{Y-X}{\sqrt{2}}$

New equation of curve becomes

$(\frac{X+Y}{\sqrt{2}})^3+(\frac{Y-X}{\sqrt{2}})^3=3a(\frac{X+Y}{\sqrt{2}})(\frac{Y-X}{\sqrt{2}})$

$\frac{1}{2\sqrt{}2}((X+Y)^3+(Y-X)^3)=(\frac{3a}{2})(Y^2-X^2)$

$(X^3+3XY^2)=\frac{3a}{\sqrt{2}}(Y^2-X^2)$

$3Y^2(X+\frac{a}{\sqrt{2}})-X^2(\frac{3a}{\sqrt{2}}-X)=0$

$3Y^2(X+b)-X^2(3b-X)=0$ where $b=\frac{a}{\sqrt{2}}$

Clearly,the asymptote of the equation parallel to Y-axis is X=-b.

Hence,the area between the curve and its asymptote is given by

$A=2\int_{-b}^{0}YdX$

$A=2\int_{-b}^{0}\frac{-x}{\sqrt{3}}(\frac{3b-x}{x+b})^{\frac{1}{2}}dX$

$A=\frac{-2}{\sqrt{3}}\int_{-b}^{0}(\frac{x(3b-x)}{\sqrt{(x+b)(3b-x)}})dX$

Putting $X=b(1-2cos\theta);dx=2bsin\theta d \theta$ for $X=-b,\theta=0;X=0,\theta=\frac{π}{3}$

$A=\frac{-2}{\sqrt{3}}\int_0^{\frac{π}{3}}\frac{b(1-2cos\theta)(3b-(-b-2bcos\theta)).2bsin\theta}{\sqrt{((b-2bcos\theta)+b)(3b-(b-2bcos\theta))}}d\theta$

$A=\frac{-2}{\sqrt{3}}\int_0^{\frac{π}{3}}2b^2(1-cos\theta-2cos^2\theta)d\theta$

$A=\frac{4b^2}{\sqrt{3}}\int_0^{\frac{π}{3}}(cos\theta+cos2\theta)d\theta$

$A=\frac{4b^2}{\sqrt{3}}[(sin\theta+\frac{sin2\theta}{2})]_0^\frac{π}{3}$

$A=3b^2=\frac{3}{2}a^2$