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, (5,6)
b. graph the tangent and normal lines on the given graph.
write the equation for the normal 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 $-2yy^\prime$. The derivative of the constant 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: Find the equation of the tangent line

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

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. So the slope of the normal line $m_{n}=-\frac{7}{16}$.

Step6: Find the equation of the normal line

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

Answer:

The equation of the tangent line is $y=\frac{16}{7}x-\frac{38}{7}$, and the equation of the normal line is $y =-\frac{7}{16}x+\frac{131}{16}$.