Question

4. Find an equation of the line that is tangent to the graph of 4x + 2xy^2 +44 = y^3 at the point (-3,2).

Answer 1

Given:

The equation is:

$4x+2xy^2+44=y^3$

To find:

The equation of the tangent on the given equation at point (-3,2).

Solution:

We have,

$4x+2xy^2+44=y^3$

Differentiate with respect to x.

$4(1)+2[x(2yy')+y^2(1)]+0=3y^2y'$

$4+4xyy'+2y^2=3y^2y'$

$4+2y^2=3y^2y'-4xyy'$

$4+2y^2=(3y^2-4xy)y'$

Isolate y'.

$\dfrac{4+2y^2}{3y^2-4xy}=y'$

Now, we need to find the value of the derivative at point (-3,2).

$y'_{(-3,2)}=\dfrac{4+2(2)^2}{3(2)^2-4(-3)(2)}$

$y'_{(-3,2)}=\dfrac{4+8}{12+24}$

$y'_{(-3,2)}=\dfrac{12}{36}$

$y'_{(-3,2)}=\dfrac{1}{3}$

It means the slope of the tangent line is $\dfrac{1}{3}$.

The equation of tangent line that passes through the point (-3,2) with slope $\dfrac{1}{3}$ is:

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

$y-2=\dfrac{1}{3}(x-(-3))$

$y-2=\dfrac{1}{3}x+1$

$y=\dfrac{1}{3}x+1+2$

$y=\dfrac{1}{3}x+3$

Therefore, the equation of tangent line is $y=\dfrac{1}{3}x+3$.