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 = 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 y = 1/5x + 26/5. b. the equation of the normal line to the curve at the point (-1,5) is

Explanation:

Step1: Differentiate the curve implicitly

Differentiate $x^{2}+y^{2}=26$ with respect to $x$. Using the chain - rule, we get $2x + 2y\frac{dy}{dx}=0$. 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 $(-1,5)$

Substitute $x = - 1$ and $y = 5$ into $\frac{dy}{dx}$. So, $m_{tangent}=\frac{dy}{dx}\big|_{x=-1,y = 5}=-\frac{-1}{5}=\frac{1}{5}$.
The equation of a line in point - slope form is $y - y_{1}=m(x - x_{1})$. Using the point $(-1,5)$ and $m=\frac{1}{5}$, we have $y - 5=\frac{1}{5}(x + 1)$, which simplifies to $y=\frac{1}{5}x+\frac{26}{5}$.

Step3: Find the slope of the normal line

The slope of the normal line $m_{normal}$ is the negative reciprocal of the slope of the tangent line. Since $m_{tangent}=\frac{1}{5}$, then $m_{normal}=-5$.

Step4: Find the equation of the normal line

Using the point - slope form $y - y_{1}=m(x - x_{1})$ with the point $(-1,5)$ and $m=-5$, we get $y - 5=-5(x + 1)$.
Expand the right - hand side: $y - 5=-5x-5$.
Add 5 to both sides to get the equation of the normal line: $y=-5x$.

Answer:

$y=-5x$