Question

verify that the given point is on the curve and find the lines that are a. tangent and b. normal to the curve at the given point.
$x^{2}+y^{2}=18,(3, - 3)$
the point is on the curve because when 3 is substituted for x and - 3 is substituted for y, the resulting statement is 18 = 18, which is a true statement.
(simplify your answers.)
a. the equation of the tangent line to the curve at the point (3, - 3) is $y=x - 6$.
b. the equation of the normal line to the curve at the point (3, - 3) is

Explanation:

Step1: Differentiate the equation implicitly

Differentiate $x^{2}+y^{2}=18$ with respect to $x$. Using the chain - rule, we have $2x + 2y\frac{dy}{dx}=0$. Then solve for $\frac{dy}{dx}$:
\[

$$\begin{align*} 2y\frac{dy}{dx}&=-2x\\ \frac{dy}{dx}&=-\frac{x}{y} \end{align*}$$

\]

Step2: Find the slope of the tangent line at the point $(3, - 3)$

Substitute $x = 3$ and $y=-3$ into $\frac{dy}{dx}$:
\[
\frac{dy}{dx}\big|_{(3,-3)}=-\frac{3}{-3}=1
\]
The equation of a line in point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(3,-3)$ and $m = 1$. So $y+3 = 1\times(x - 3)$, which simplifies to $y=x - 6$.

Step3: 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 $m_{t}=1$, the slope of the normal line $m_{n}=-1$.

Step4: Find the equation of the normal line

Using the point - slope form $y - y_{1}=m(x - x_{1})$ with $(x_{1},y_{1})=(3,-3)$ and $m=-1$, we get $y+3=-1\times(x - 3)$.
\[

$$\begin{align*} y+3&=-x + 3\\ y&=-x \end{align*}$$

\]

Answer:

$y=-x$