Question

a. determine an equation of the tangent line and the normal line at the given point on the curve. x^2+xy - y^2=19,(4,1)
b. graph the tangent and normal lines on the given graph.
a. write the equation for the tangent line.

Explanation:

Step1: Differentiate implicitly

Differentiate $x^{2}+xy - y^{2}=19$ with respect to $x$.
Using the sum - rule and product - rule, we have:
$\frac{d}{dx}(x^{2})+\frac{d}{dx}(xy)-\frac{d}{dx}(y^{2})=\frac{d}{dx}(19)$.
$2x + y+x\frac{dy}{dx}-2y\frac{dy}{dx}=0$.

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

Rearrange the terms to isolate $\frac{dy}{dx}$:
$x\frac{dy}{dx}-2y\frac{dy}{dx}=-2x - y$.
$\frac{dy}{dx}(x - 2y)=-2x - y$.
$\frac{dy}{dx}=\frac{-2x - y}{x - 2y}$.

Step3: Find the slope of the tangent at the point $(4,1)$

Substitute $x = 4$ and $y = 1$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}\big|_{(4,1)}=\frac{-2\times4-1}{4-2\times1}=\frac{-8 - 1}{4 - 2}=\frac{-9}{2}$.

Step4: Write the equation of the tangent line

Use the point - slope form $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(4,1)$ and $m =-\frac{9}{2}$.
$y - 1=-\frac{9}{2}(x - 4)$.
$y-1=-\frac{9}{2}x + 18$.
$y=-\frac{9}{2}x+19$.

Step5: Find the slope of the normal line

The slope of the normal line is the negative reciprocal of the slope of the tangent line.
If the slope of the tangent line is $m_{t}=-\frac{9}{2}$, then the slope of the normal line $m_{n}=\frac{2}{9}$.

Step6: Write the equation of the normal line

Use the point - slope form $y - y_{1}=m(x - x_{1})$ with $(x_{1},y_{1})=(4,1)$ and $m=\frac{2}{9}$.
$y - 1=\frac{2}{9}(x - 4)$.
$y-1=\frac{2}{9}x-\frac{8}{9}$.
$y=\frac{2}{9}x+\frac{1}{9}$.

Answer:

Tangent line: $y =-\frac{9}{2}x + 19$
Normal line: $y=\frac{2}{9}x+\frac{1}{9}$