Question

find all points on the curve where it has a horizontal tangent line

Answer 1

A horizontal tangent line is a mathematical feature on a graph, located where a function's derivative is zero.

Hence, to find the points on the curve where it has a horizontal tangent line, Find the derivative of the given equation of curve and equate it to zero.

The given equation of curve is:

$3xy^2-18x^2y=8$

Find the derivative using implicit differentiation:

$\begin{gathered} \frac{d}{dx}(3xy^2)-\frac{d}{dx}(18x^2y)=\frac{d}{dx}(8) \\ Use\text{ the scalar product property and derivative of constant property of derivatives:} \\ \Rightarrow3\frac{d}{dx}(xy^2)-18\frac{d}{dx}(x^2y)=0 \\ \Rightarrow3(y^2\frac{d}{dx}x+x\frac{d}{dx}y^2)-18(y\frac{d}{dx}x^2+x^2\frac{dy}{dx})=0 \\ \Rightarrow3(y^2\times1+2xy\frac{dy}{dx})-18(2xy+x^2\frac{dy}{dx})=0 \\ \Rightarrow3y^2+6xy\frac{dy}{dx}-36xy-18x^2\frac{dy}{dx}=0 \\ \Rightarrow6xy\frac{dy}{dx}-18x^2\frac{dy}{dx}=-3y^2+36xy \\ \Rightarrow\frac{dy}{dx}(6xy-18x^2)=-3y^2+36xy \\ \Rightarrow\frac{dy}{dx}=\frac{-3y^2+36xy}{6xy-18x^2} \end{gathered}$

Equate the derivative to zero:

$\begin{gathered} \frac{-3y^2+36xy}{6xy-18x^2}=0 \\ \Rightarrow-3y^2+36xy=0 \end{gathered}$

Solve the equation:

$\begin{gathered} -3y^2+36xy=0 \\ \Rightarrow-3y(y-12x)=0 \\ Divide\text{ both sides by -3:} \\ \Rightarrow y(y-12x)=0 \\ \Rightarrow y=0\text{ or }y-12x=0 \\ \Rightarrow y=0\text{ or }y=12x \end{gathered}$

The solution y=0 does not satisfy the equation of the curve. Hence discard it.

We need to find points on the curve where y=12x.

Hence substitute y=12x into the equation of the curve:

$\begin{gathered} 3xy^2-18x^2y=8;y=12x \\ \Rightarrow3x(12x)^2-18x^2(12x)=8 \\ \Rightarrow3x(144x^2)-216x^3=8 \\ \Rightarrow432x^3-216x^3=8 \\ \Rightarrow216x^3=8 \\ \Rightarrow x^3=\frac{8}{216} \\ \Rightarrow x^3=(\frac{2}{6})^3 \\ \Rightarrow x=\frac{2}{6}=\frac{1}{3} \end{gathered}$

Substitute x=2/6 into the equation of curve:

$\begin{gathered} 3xy^2-18x^2y=8;x=\frac{1}{3} \\ \Rightarrow3(\frac{1}{3})y^2-18(\frac{1}{3})^2y=8 \\ \Rightarrow y^2-2y=8 \\ \Rightarrow y^2-2y-8=0 \\ \Rightarrow y^2-4y+2y-8=0 \\ \Rightarrow y(y-4)+2(y-4)=0 \\ \Rightarrow(y+2)(y-4)=0 \\ \Rightarrow y=-2,4 \end{gathered}$

Notice that y=-2 does not satisfy the equation y=12x for x=1/3.

It follows that the required point is only:

$(\frac{1}{3},4)$

The required point of horizontal tangent line is (1/3, 4).