Question

which of the following is an equation of the line tangent to the graph of $x^{2}-3xy = 10$ at the point $(1,-3)$?
(a) $y + 3=-11(x - 1)$
(b) $y + 3=-\frac{1}{3}(x - 1)$
(c) $y + 3=\frac{1}{3}(x - 1)$
(d) $y + 3=\frac{7}{3}(x - 1)$
(e) $y+3=\frac{11}{3}(x - 1)$

Explanation:

Step1: Differentiate implicitly

Differentiate $x^{2}-3xy = 10$ with respect to $x$. Using the power - rule and product - rule. The derivative of $x^{2}$ is $2x$. For $-3xy$, by the product rule $(uv)^\prime=u^\prime v + uv^\prime$ where $u=-3x$ and $v = y$, we have $-3y-3x\frac{dy}{dx}$. The derivative of the constant $10$ is $0$. So, $2x-3y - 3x\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Rearrange the equation $2x-3y - 3x\frac{dy}{dx}=0$ to isolate $\frac{dy}{dx}$. First, move the non - $\frac{dy}{dx}$ terms to the other side: $-3x\frac{dy}{dx}=3y - 2x$. Then, $\frac{dy}{dx}=\frac{2x - 3y}{3x}$.

Step3: Evaluate $\frac{dy}{dx}$ at the point $(1,-3)$

Substitute $x = 1$ and $y=-3$ into $\frac{dy}{dx}=\frac{2x - 3y}{3x}$. We get $\frac{2(1)-3(-3)}{3(1)}=\frac{2 + 9}{3}=\frac{11}{3}$.

Step4: Use the point - slope form of a line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,-3)$ and $m=\frac{11}{3}$. Substituting these values, we get $y+3=\frac{11}{3}(x - 1)$.

Answer:

E. $y + 3=\frac{11}{3}(x - 1)$