Answered

write the equation for the tangent line in slope-intercept form.

2x^2+xy+y^2=36; (3,3)

Verified Answer :

\(\bf 2x^2+xy+y^2=36\\\\ -----------------------------\\\\ 4x+\left[1\cdot y+x\frac{dy}{dx} \right]+2y\frac{dy}{dx}=0 \\\\\\ \cfrac{dy}{dx}(x+2y)=-4x-y\implies \left. \cfrac{dy}{dx}=\cfrac{-4x-y}{x+2y} \right|_{(3,3)}\implies -\cfrac{5}{3}\leftarrow m\\\\ -----------------------------\\\\ y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-3=-\cfrac{5}{3}(x-3)\\ \qquad \uparrow\\ \textit{point-slope form}\)

and solve that for "y", that'd be the equation for the tangent line at 3,3

Related Questions