The following equation defines a family of plane curves. x3 + y3 - 3axy = c wher

Question

The following equation defines a family of plane curves. x3 + y3 - 3axy = c where c ( - infinity; + infinity). A unique family member passes through every point in the XY-plane except (0. 0) and (a, a). Let (alpha, beta) be a point on one of these plane curves. Your particular parameter and coordinates are a = -8 and (alpha, beta) = (-4,-8). Find the gradient of your curve (the value of its derivative) at your given point (alpha, beta) = (-4, -8). Find an explicit equation for the tangent at the point (beta, alpha) = (-4,-8). Show or explain why the reversed coordinate point (beta, alpha) = (-8,-4) is also a point on x3 + y3 -3 axy = c. Find the gradient of the curve at the point (beta, alpha) = (-8, -4)

Explanation / Answer

the curve will pass thru (1,0) this gives 1-0-0=c c=1 it will also pass thru (1,1) this gives 1+1-3a=1 a=1/3 so basically curve is x^3+Y^3-XY=1 GRADIENT OF CURVE= 3X^2+3Y^2DY/DX-XDY/DX-Y=0 AT POINT(-4,-8) 3*16+3*64DY/DX+4DY/DX+8=0 DY/DX=-(56/196)=-14/49

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