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 = 5; (3,4)
b. graph the tangent and normal lines on the given graph.
a. write the equation for the tangent line.

Explanation:

Step1: Differentiate implicitly

Differentiate $x^{2}+xy - y^{2}=5$ with respect to $x$.
Using the sum - rule and product - rule:
The derivative of $x^{2}$ is $2x$. For $xy$, by the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x$ and $v = y$, we get $y + x\frac{dy}{dx}$. The derivative of $-y^{2}$ is $-2y\frac{dy}{dx}$, and the derivative of the constant 5 is 0.
So, $2x + y+x\frac{dy}{dx}-2y\frac{dy}{dx}=0$.

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

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

Step3: Find the slope of the tangent at the point $(3,4)$

Substitute $x = 3$ and $y = 4$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}\big|_{(3,4)}=\frac{-2\times3 - 4}{3-2\times4}=\frac{-6 - 4}{3 - 8}=\frac{-10}{-5}=2$.

Step4: Find the equation of the tangent line

Use the point - slope form of a line $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(3,4)$ and $m = 2$.
$y - 4=2(x - 3)$.
Expand: $y-4 = 2x-6$.
The equation of the tangent line is $y=2x - 2$.

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. Since the slope of the tangent line is 2, the slope of the normal line is $-\frac{1}{2}$.

Step6: Find the equation of the normal line

Use the point - slope form $y - y_{1}=m(x - x_{1})$ with $(x_{1},y_{1})=(3,4)$ and $m=-\frac{1}{2}$.
$y - 4=-\frac{1}{2}(x - 3)$.
Expand: $y-4=-\frac{1}{2}x+\frac{3}{2}$.
$y=-\frac{1}{2}x+\frac{3}{2}+4=-\frac{1}{2}x+\frac{3 + 8}{2}=-\frac{1}{2}x+\frac{11}{2}$.

Answer:

The equation of the tangent line is $y = 2x-2$.
The equation of the normal line is $y=-\frac{1}{2}x+\frac{11}{2}$.