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.
$x^{2}+y^{2}=26,(-1,5)$
the point is on the curve because when - 1 is substituted for x and 5 is substituted for y, the resulting statement is 26 = 26, which is a true statement.
(simplify your answers.)
a. the equation of the tangent line to the curve at the point (-1,5) is

Explanation:

Step1: Differentiate implicitly

Differentiate $x^{2}+y^{2}=26$ with respect to $x$. Using the chain - rule, we have $2x + 2y\frac{dy}{dx}=0$.

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 $x = - 1$ and $y = 5$ into $\frac{dy}{dx}$:
\[
m_{tangent}=-\frac{-1}{5}=\frac{1}{5}
\]

Step4: Use the point - slope form for the tangent line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(-1,5)$ and $m=\frac{1}{5}$.
\[

$$\begin{align*} y - 5&=\frac{1}{5}(x + 1)\\ y-5&=\frac{1}{5}x+\frac{1}{5}\\ y&=\frac{1}{5}x+\frac{1}{5}+5\\ y&=\frac{1}{5}x+\frac{26}{5} \end{align*}$$

\]

Answer:

$y=\frac{1}{5}x+\frac{26}{5}$