Answer to Question #180047 in Calculus for SAHIL

Question

The tangents at the origin of the curve x^3+y^3=3axy are

Answer 1

To find the tangent to curve at the origin,we need to equate the lowest degree term to 0

Therefore, 3axy=0, which gives x=0, and y =0, as two tangent to the curve at the origin.

Point(3a2,3a2) and also point (0, 0) i.e origin are exactly the points of intersection of line y=x

y=x with curve (x^3 + y^3)= 3axy

(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

y=x, that is the gradient of tangent lines is m =−1

m=−1 which means its angle with positive direction of x axis is obtuse.The equation of tangent line is:

y−32=−1(x−32)

y−32=−1(x−32)

Or: y=−x+3

Differentiate (x^3)+ (y^3)−3axy=0

(x^3 + y^3−3axy)=0 implicitly and solve for the slope

′y = y′(x)=(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 3π/4

3π/4 obtuse angle to x-axis.