Question

Find equation of the tangent to x2 +xy + 2y3 = 4 at (– 2, 1).

Answer 1

Given:

$x^2+xy+2y^3=4$

Let's find the equation of the tangent to the given equation at the point:

(x, y) ==> (-2, 1)

Let's first find the derivative of the given equation.

We have:

$\begin{gathered} \frac{d}{dx}(x^2+xy+2y^3)=\frac{d}{dx}(4) \\ \\ 2x+y+xy^{\prime}+6y^2y^{\prime}=0 \end{gathered}$

Now, rewrite the equation for y':

$\begin{gathered} xy^{\prime}+6y^2y^{\prime}=-2x-y \\ \\ \text{ Factor out y':} \\ y^{\prime}(x+6y^2)=-2x-y \\ \\ \text{ Divide both sides by \lparen x}+6y^2): \\ \frac{y^{\prime}(x+6y^2)}{x+6y^2}=\frac{-2x-y}{x+6y^2} \\ \\ y^{\prime}=\frac{-2x-y}{x+6y^2} \\ \\ y^{\prime}=-\frac{2x+y}{x+6y^2} \end{gathered}$

Now, to find the slope of the tangent, evaluate the derivative at the point (-2, 1).

Substitute -2 for x and 1 for y:

$\begin{gathered} m=-\frac{2(-2)+1}{-2+6(1)^2} \\ \\ m=-\frac{-4+1}{-2+6} \\ \\ m=-\frac{-3}{4} \\ \\ m=\frac{3}{4} \end{gathered}$

The slope of the tangent line is 3/4.

Now, to find the equation of the tangent line, apply the point-slope form:

$y-y_1=m(x-x_1)$

Plug in -2 for x1 and 1 for y1 and 3/4 for m:

$\begin{gathered} y-1=\frac{3}{4}(x-(-2)) \\ \\ y-1=\frac{3}{4}(x+2) \\ \\ y-1=\frac{3}{4}x+\frac{3}{4}*2 \\ \\ y-1=\frac{3}{4}x+\frac{3*2}{4} \end{gathered}$

Solving further:

$\begin{gathered} y-1=\frac{3}{4}x+\frac{6}{4} \\ \\ y-1=\frac{3}{4}x+\frac{3}{2} \\ \\ Add\text{ 1 to both sides:} \\ y-1+1=\frac{3}{4}x+\frac{3}{2}+1 \\ \\ y=\frac{3}{4}x+\frac{3+2}{2} \\ \\ y=\frac{3}{4}x+\frac{5}{2} \end{gathered}$

Therefore, the equation of the tangent line is:

$y=\frac{3}{4}x+\frac{5}{2}$

• ANSWER:

$y=\frac{3}{4}x+\frac{5}{2}$