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)

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:
The derivative of $4x^{2}$ is $8x$.
For $4xy$, by the product - rule $(uv)^\prime=u^\prime v+uv^\prime$ where $u = 4x$ and $v = y$, we get $4y+4x\frac{dy}{dx}$.
The derivative of $6y^{2}$ using the chain - rule is $12y\frac{dy}{dx}$.
The derivative of the right - hand side (a constant 14) is 0.
So, $8x + 4y+4x\frac{dy}{dx}+12y\frac{dy}{dx}=0$.

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

Group the terms with $\frac{dy}{dx}$ together:
$4x\frac{dy}{dx}+12y\frac{dy}{dx}=-8x - 4y$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(4x + 12y)=-8x - 4y$.
Then $\frac{dy}{dx}=\frac{-8x - 4y}{4x + 12y}=\frac{-2x - y}{x + 3y}$.

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

Substitute $x = 1$ and $y = 1$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}\big|_{(1,1)}=\frac{-2(1)-1}{1 + 3(1)}=\frac{-2 - 1}{1+3}=-\frac{3}{4}$.

Step4: Find the slope of the normal line

The slope of the normal line $m_n$ and the slope of the tangent line $m_t$ are related by $m_n=-\frac{1}{m_t}$.
Since $m_t=-\frac{3}{4}$, then $m_n=\frac{4}{3}$.

Step5: Find the equation of the normal line

Use the point - slope form of a line $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(1,1)$ and $m=\frac{4}{3}$.
$y - 1=\frac{4}{3}(x - 1)$.
Expand: $y-1=\frac{4}{3}x-\frac{4}{3}$.
$y=\frac{4}{3}x-\frac{4}{3}+1=\frac{4}{3}x-\frac{1}{3}$.
The equation of the normal line is $4x-3y = 1$.

Answer:

The equation of the normal line to the curve $4x^{2}+4xy + 6y^{2}=14$ at the point $(1,1)$ is $4x - 3y=1$.