prove : The tangent to the curve ${x}^3+{y}^3=3axy$ at $(\frac{3a}{2},\frac{3a}{2})$ makes obtuse angle with the positive direction of $x$ axis?

Question

I have differentiated the equation from both side and put the value of coordinates given. Is there any better way? Can anyone help me to solve the problem?

Answer 1

Hint:

Point $(\frac {3a}2, \frac{3a}2)$ and also point (0, 0) i.e origin are exactly the points of intersection of line $y=x$ with curve $x^3+y^3=3axy$. if you plot the curve and line you see that the tangent line at that point is perpendicular on line $y=x$, that is the gradient of tangent lines is $m=-1$ which means its angle with positive direction of x axis is obtuse.The equation of tangent line is:

$y-\frac 32=-1(x-\frac 32)$

Or: $y=-x +3$

Answer 2

I trust that you tried. Only the last step did not connect somewhere.

Differentiate $ x^3+y^3-3 a x y =0$ implicitly and solve for the slope

$$ y'=y'(x)=\frac{ay-x^2}{y^2-ax}$$

The given point has

$$x=y$$

Plug in and the slope reduces to $-1$, so the the tangent makes CCW $ 3 \pi/4$ obtuse angle to x-axis.