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} (d)/(dx)(3xy^2)-(d)/(dx)(18x^2y)=(d)/(dx)(8) \\ Use\text{ the scalar product property and derivative of constant property of derivatives:} \\ \Rightarrow3(d)/(dx)(xy^2)-18(d)/(dx)(x^2y)=0 \\ \Rightarrow3(y^2(d)/(dx)x+x(d)/(dx)y^2)-18(y(d)/(dx)x^2+x^2(dy)/(dx))=0 \\ \Rightarrow3(y^2*1+2xy(dy)/(dx))-18(2xy+x^2(dy)/(dx))=0 \\ \Rightarrow3y^2+6xy(dy)/(dx)-36xy-18x^2(dy)/(dx)=0 \\ \Rightarrow6xy(dy)/(dx)-18x^2(dy)/(dx)=-3y^2+36xy \\ \Rightarrow(dy)/(dx)(6xy-18x^2)=-3y^2+36xy \\ \Rightarrow(dy)/(dx)=(-3y^2+36xy)/(6xy-18x^2) \end{gathered}`

Equate the derivative to zero:

`\begin{gathered} (-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=(8)/(216) \\ \Rightarrow x^3=((2)/(6))^3 \\ \Rightarrow x=(2)/(6)=(1)/(3) \end{gathered}`

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

`\begin{gathered} 3xy^2-18x^2y=8;x=(1)/(3) \\ \Rightarrow3((1)/(3))y^2-18((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:

`((1)/(3),4)`

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