use implicit differentiation to find an equation of the tangent line to the curve at the given point. x^2…

Explanation:

Step1: Differentiate both sides

Differentiate $x^{2}-xy - y^{2}=1$ with respect to $x$. Using the power - rule and product - rule: The derivative of $x^{2}$ is $2x$. For $-xy$, by the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u=-x$ and $v = y$, we have $-(y + xy^\prime)$. The derivative of $-y^{2}$ is $-2yy^\prime$ and the derivative of 1 is 0. So, $2x-(y + xy^\prime)-2yy^\prime=0$.

Step2: Solve for $y^\prime$

Expand the left - hand side: $2x - y-xy^\prime-2yy^\prime = 0$. Group the terms with $y^\prime$: $-xy^\prime-2yy^\prime=y - 2x$. Factor out $y^\prime$: $y^\prime(-x - 2y)=y - 2x$. Then $y^\prime=\frac{2x - y}{x + 2y}$.

Step3: Find the slope at the given point

Substitute $x = 5$ and $y = 3$ into $y^\prime$: $y^\prime=\frac{2\times5-3}{5 + 2\times3}=\frac{10 - 3}{5+6}=\frac{7}{11}$.

Step4: Use the point - slope form

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(5,3)$ and $m=\frac{7}{11}$. $y - 3=\frac{7}{11}(x - 5)$. Multiply through by 11 to get $11y-33 = 7x-35$. Rearrange to the general form: $7x-11y=2$.

Answer:

$7x - 11y=2$