Science:Math Exam Resources/Courses/MATH104/December 2011/Question 01 (i)/Solution 1

To determine the equation of the tangent line we will need the slope of the line at the given point. To find the slope at the given point, we may use implicit differentiation and find an expression for y'.

${\begin{aligned}{\frac {d}{dx}}{\Big (}xy^{2}+2xy{\Big )}&={\frac {d}{dx}}(8)\\y^{2}+2xyy'+2y+2xy'&=0\\\end{aligned}}$

At the point (1,2), this equation becomes

${\begin{aligned}(2)^{2}+2(1)(2)y'+2(2)+2(1)y'&=0\\4+4y'+4+2y'&=0\\6y'&=-8\\y'&=-4/3\end{aligned}}$

Using the "point-slope" equation of a straight line with slope m = y' that passes through the point (x₀,y₀)

$\displaystyle y-y_{0}=m(x-x_{0})$

we have

${\begin{aligned}y-(2)&=(-{\frac {4}{3}}){\big (}x-(1){\big )}\\y&=-{\frac {4}{3}}x+{\frac {10}{3}}\\\end{aligned}}$

which is the equation of the tangent line at (1,2).