Re: John's question at Yahoo! Answers regarding implicit differentiation & horizontal/vertical tange
Hello John,
We are given the equation:
$$x^2+xy+y^2=y$$
Implicitly differentiating with respect to $x$, we find:
$$2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=\frac{dy}{dx}$$
Solving for $$\frac{dy}{dx}$$, we obtain:
$$x\frac{dy}{dx}+2y\frac{dy}{dx}-\frac{dy}{dx}=-2x-y$$
$$\frac{dy}{dx}(x+2y-1)=-2x-y$$
$$\frac{dy}{dx}=\frac{2x+y}{1-x-2y}$$
Now, the tangent line(s) will be horizontal where the numerator is zero, which implies:
$$y=-2x$$
Substituting this into the original equation, we obtain:
$$x^2+x(-2x)+(-2x)^2=(-2x)$$
$$x^2-2x^2+4x^2=-2x$$
$$3x^2+2x=0$$
$$x(3x+2)=0$$
$$x=0,\,-\frac{2}{3}$$
Hence we have two points from $(x,-2x)$:
$$(0,0),\,\left(-\frac{2}{3},\frac{4}{3} \right)$$
Here is a plot of the curve with its two horizontal tangent lines:
View attachment 1481
The tangent line(s) will be vertical where the denominator is zero, which implies:
$$y=\frac{1-x}{2}$$
Substituting this into the original equation, we obtain:
$$x^2+x\left(\frac{1-x}{2} \right)+\left(\frac{1-x}{2} \right)^2=\left(\frac{1-x}{2} \right)$$
$$4x^2+2x(1-x)+(1-x)^2=2(1-x)$$
$$4x^2+2x-2x^2+1-2x+x^2=2-2x$$
$$3x^2+2x-1=0$$
$$(3x-1)(x+1)=0$$
$$x=-1,\,\frac{1}{3}$$
Hence we have two points from $$\left(x,\frac{1-x}{2} \right)$$:
$$\left(-1,1 \right),\,\left(\frac{1}{3},\frac{1}{3} \right)$$
Here is a plot of the curve with its two vertical tangent lines:
View attachment 1482
