Question

a normal line to a curve passes through a point p on the curve perpendicular to the line tangent to the curve at p (see figure). use the following equation and graph to determine an equation of the normal line at the given point and illustrate your work by graphing the curve with the normal line. 4x^{2}+4xy + 6y^{2}=14;(1,1) write the equation of the normal line at the given point. y = □

Explanation:

Step1: Differentiate the given equation implicitly

Differentiate $4x^{2}+4xy + 6y^{2}=14$ with respect to $x$.
Using the sum - rule and product - rule, we have:
\[

$$\begin{align*} \frac{d}{dx}(4x^{2})+\frac{d}{dx}(4xy)+\frac{d}{dx}(6y^{2})&=\frac{d}{dx}(14)\\ 8x + 4y+4x\frac{dy}{dx}+12y\frac{dy}{dx}&=0 \end{align*}$$

\]

Step2: Solve for $\frac{dy}{dx}$

\[

$$\begin{align*} 4x\frac{dy}{dx}+12y\frac{dy}{dx}&=- 8x - 4y\\ \frac{dy}{dx}(4x + 12y)&=-8x - 4y\\ \frac{dy}{dx}&=\frac{-8x - 4y}{4x + 12y}=\frac{-2x - y}{x + 3y} \end{align*}$$

\]

Step3: Find the slope of the tangent at the point $(1,1)$

Substitute $x = 1$ and $y = 1$ into $\frac{dy}{dx}$:
\[

$$\begin{align*} m_{tangent}&=\frac{-2(1)-1}{1 + 3(1)}\\ &=\frac{-2 - 1}{1+3}\\ &=-\frac{3}{4} \end{align*}$$

\]

Step4: Find the slope of the normal

Since the slope of the normal $m_{normal}$ and the slope of the tangent $m_{tangent}$ are negative reciprocals of each other, $m_{normal}=\frac{4}{3}$ (because $m_{tangent}\times m_{normal}=- 1$).

Step5: Use the point - slope form to find the equation of the normal

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,1)$ and $m=\frac{4}{3}$.
\[

$$\begin{align*} y - 1&=\frac{4}{3}(x - 1)\\ y-1&=\frac{4}{3}x-\frac{4}{3}\\ y&=\frac{4}{3}x-\frac{4}{3}+1\\ y&=\frac{4}{3}x-\frac{1}{3} \end{align*}$$

\]

Answer:

$y=\frac{4}{3}x-\frac{1}{3}$