find the equation of the tangent line at the given point on the following curve. x² + y² = 26, (1,5) the…

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 at the given point

Substitute $x = 1$ and $y = 5$ into $\frac{dy}{dx}$: $\frac{dy}{dx}\big|_{(1,5)}=-\frac{1}{5}$.

Step4: Use the point - slope form

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:

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