verify that the given point is on the curve and find the lines that are a. tangent and b. normal to the…

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.\n$x^{2}+y^{2}=26,(-1,5)$\nthe 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.\n(simplify your answers.)\n a. the equation of the tangent line to the curve at the point (-1,5) is

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.\n$x^{2}+y^{2}=26,(-1,5)$\nthe 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.\n(simplify your answers.)\n a. the equation of the tangent line to the curve at the point (-1,5) is

Answer

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:

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