Question

Find the equation of the line tangent to the curve x³ + y³ = 2xy at (1,1).

Answer 1

`y = -x +2`

Explanation:

`\text{Given that,}\\\\~~~~~~~~~~x^3 +y^3 = 2xy\\\\\implies (d)/(dx)(x^3 +y^3) = 2(d)/(dx)(xy)\\\\\implies 3x^2 +3y^2 (dy)/(dx) = 2 \left( y + x(dy)/(dx)\right)\\\\\implies 3x^2 + 3y^2 (dy)/(dx) = 2y + 2x(dy)/(dx)\\\\\implies 3y^2 (dy)/(dx)-2x (dy)/(dx) = 2y - 3x^2\\\\\implies (3y^2 -2x) (dy)/(dx) = 2y-3x^2\\\\\implies (dy)/(dx) = (2y-3x^2)/(3y^2 -2x)\\`

`\text{At (1,1),}\\\\\text{Slope,}~ m = (dy)/(dx)\\\\~~~~~~~~~~~~~=(2(1) -3(1^2))/(3(1^2) -2(1))\\\\~~~~~~~~~~~~~=(2-3)/(3-2)\\\\~~~~~~~~~~~~~=(-1)/(1)\\\\~~~~~~~~~~~~~=-1`

`\text{Equation of tangent line,}\\\\~~~~~~~~~y -y_1 = m(x-x_1)\\\\\implies y -1 = -1(x-1)\\\\\implies y -1 = -x+1\\\\\implies y = -x +1+1\\\\\implies y = -x +2`