Question

Calculus Questions using implicit differentiation to find the equation of the tangent line that is given curve at the 2, -1

Answer 1

The tangent line to the curve has slope equal to $\frac{\mathrm dy}{\mathrm dx}$ at the point (2, -1). We have, by implicit differentiation,

$3y^2-6xy=2x+11\implies6y\dfrac{\mathrm dy}{\mathrm dx}-6y-6x\dfrac{\mathrm dy}{\mathrm dx}=2$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2+6y}{6y-6x}=\dfrac{1+3y}{3y-3x}$

At the point (2, -1), the slope is then 2/9, so the tangent line has equation

$y-(-1)=\dfrac29(x-2)\implies y=\dfrac29x-\dfrac{13}9$