Question

I need help with part b.Answer of Part A: (dy(x))/(dx) = (2 x - sqrt(3) y)/(sqrt(3) x - 4 y)

Answer 1

Given the equation:

`x^2-\surd3xy+2y^2=12`

You know that the derivative with respect to "x" is:

`(dy)/(dx)=\frac{2x-\sqrt[]{3}y}{\sqrt[]{3}x-4y}`

• The Slope-Intercept Form of the equation of a line is:

`y=mx+b`

Where "m" is the slope and "b" is the y-intercept.

In this case, to find the slope of the tangent line, you need to substitute the coordinates of the point given in the exercise, into the derivative found in Part A. Then, knowing that:

`\begin{gathered} x=\sqrt[]{3} \\ y=3 \end{gathered}`

You get:

`m=\frac{2(\sqrt[]{3})-\sqrt[]{3}(3)}{\sqrt[]{3}(\sqrt[]{3})-4(3)}=\frac{2\sqrt[]{3}-3\sqrt[]{3}}{3-12}=\frac{-\sqrt[]{3}}{-9}=\frac{\sqrt[]{3}}{9}`

Substitute the slope and the coordinates of the point into this equation:

`y=mx+b`

Then:

`3=(\frac{\sqrt[]{3}}{9})(\sqrt[]{3})+b`

Solve for "b":

`\begin{gathered} 3=(\frac{\sqrt[]{3}}{9})(\sqrt[]{3})+b \\ \\ 3=\frac{(\sqrt[]{3})^2}{9}+b \\ \\ 3=(3)/(9)+b \end{gathered}`
`\begin{gathered} 3-(3)/(9)=b \\ \\ b=(8)/(3) \end{gathered}`

Therefore, you can write this equation of the tangent line:

`y=\frac{\sqrt[]{3}}{9}x+(8)/(3)`

• In order to find the equation of the line normal line at the same point, you need to

remember that it is perpendicular to the tangent line.

By definition, the slopes of perpendicular lines are opposite reciprocal. Therefore, knowing the slope of the tangent line, you can determine that the slope of the normal line is:

`m_{\text{normal}}=-\frac{9}{\sqrt[]{3}}`

You can simplify as follows:

`m_{\text{normal}}=-\frac{9}{\sqrt[]{3}}\cdot\frac{\sqrt[]{3}}{\sqrt[]{3}}`
`\begin{gathered} m_{\text{normal}}=-\frac{9\sqrt[]{3}}{3} \\ \\ m_{\text{normal}}=-3\sqrt[]{3} \end{gathered}`

Knowing the point given in the exercise and the slope of the normal line, you can substitute them into this equation and solve for "b":

`y=mx+b`

Then, you get:

`\begin{gathered} 3=(-3\sqrt[]{3)}(\sqrt[]{3})+b \\ \\ 3=-3(\sqrt[]{3)}^2+b \end{gathered}`
`\begin{gathered} 3=-3(3)+b \\ 3+9=b \\ b=12 \end{gathered}`

Finally, knowing the slope and the y-intercept, you can write the equation of the normal line:

`y=-3\sqrt[]{3}x+12`

Hence, the answer is:

• Equation of the tangent line:

`y=\frac{\sqrt[]{3}}{9}x+(8)/(3)`

• Equation of the normal line:

`y=-3\sqrt[]{3}x+12`