Question

a. determine an equation for the tangent line and the normal line at the given point on the curve. x²+xy - y²=19, (5,6)
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:
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 $y+xy^\prime$. The derivative of $-y^{2}$ is $-2yy^\prime$ and the derivative of 19 is 0.
So, $2x + y+xy^\prime-2yy^\prime = 0$.

Step2: Solve for $y^\prime$

Rearrange the terms to isolate $y^\prime$:
$xy^\prime-2yy^\prime=-2x - y$.
Factor out $y^\prime$: $y^\prime(x - 2y)=-2x - y$.
Then $y^\prime=\frac{-2x - y}{x - 2y}$.

Step3: Find the slope of the tangent line at the point $(5,6)$

Substitute $x = 5$ and $y = 6$ into $y^\prime$:
$y^\prime=\frac{-2\times5-6}{5 - 2\times6}=\frac{-10 - 6}{5-12}=\frac{-16}{-7}=\frac{16}{7}$.

Step4: Write the equation of the tangent line

Use the point - slope form $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(5,6)$ and $m=\frac{16}{7}$.
$y - 6=\frac{16}{7}(x - 5)$.
$7y-42 = 16x-80$.
$16x-7y=38$.

Step5: Find the slope of the normal line

The slope of the normal line $m_{n}$ is the negative reciprocal of the slope of the tangent line. So $m_{n}=-\frac{7}{16}$.

Step6: Write the equation of the normal line

Using the point - slope form $y - y_{1}=m_{n}(x - x_{1})$ with $(x_{1},y_{1})=(5,6)$ and $m_{n}=-\frac{7}{16}$.
$y - 6=-\frac{7}{16}(x - 5)$.
$16y-96=-7x + 35$.
$7x+16y=131$.

Answer:

Equation of the tangent line: $16x - 7y=38$
Equation of the normal line: $7x + 16y=131$