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. x² + y² = 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

Explanation:

Step1: Differentiate the equation implicitly

Differentiate $x^{2}+y^{2}=18$ with respect to $x$. Using the power - rule and chain - rule, we get $2x + 2y\frac{dy}{dx}=0$.

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

Isolate $\frac{dy}{dx}$:
\[

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

\]

Step3: Find the slope of the tangent line

Substitute $x = 3$ and $y=-3$ into $\frac{dy}{dx}$:
\[
\frac{dy}{dx}\big|_{(3,-3)}=-\frac{3}{-3}=1
\]

Step4: Use the point - slope form to find the tangent line equation

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(3,-3)$ and $m = 1$.
\[

$$\begin{align*} y-(-3)&=1\times(x - 3)\\ y + 3&=x-3\\ y&=x-6 \end{align*}$$

\]

Answer:

$y=x - 6$