Question

Write the equation of the line tangent to $x^{2} y+y^3+4x=8$ at the point where the given curve intersects the x-axis.

Answer 1

yay, calc

we need to take take the implicit derivitive
to evaluate the slope, we need the point

when the line crosses the x axis, the value of y is 0
find the x value when y=0 to find the point, (x,0)

0+0+4x=8
4x=8
x=2
(2,0) Is the point



so

the derivitive is
$2xy+x^2 \frac{dy}{dx}+3y^2 \frac{dy}{dx}+4=0$
minus 2xy both sides
$x^2 \frac{dy}{dx}+3y^2 \frac{dy}{dx}+4=-2xy$
minus 4 both sides
$x^2 \frac{dy}{dx}+3y^2 \frac{dy}{dx}=-2xy-4$
undistribute dy/dx
$\frac{dy}{dx}(x^2+3y^2)=-2xy-4$
divide both sides by x²+3y²
$\frac{dy}{dx} = \frac{-2xy-4}{x^2+3y^2}$
now find the slope at the point (2,0)

sub 2 for x and 0 for y
$\frac{dy}{dx} = \frac{-2(2)(0)-4}{2^2+3(0)^2}$
$\frac{dy}{dx} = \frac{0-4}{4+0}$
$\frac{dy}{dx} = \frac{-4}{4}$
$\frac{dy}{dx} = -1$
slope is -1


use point slope form
the equation of a line that has a slope of m and passes through the point (x1,y1) is y-y1=m(x-x1)
slope is -1 and point is (2,0)

y-0=-1(x-2)
y=-1(x-2)
or
y=-x+2