RaySix

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at October 31st 2023, 6:30:16 pm.

Consider the equation that implicitly defines a curve:

x^2 + 2xy - y^2 = 12

(a)

(b) Find the equation of the tangent line to the curve at the point $(4, -2)$ .

(a) To find $\frac{{dy}}{{dx}}$ , we differentiate both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ .

Differentiating the left-hand side:

\frac{{d}}{{dx}}(x^2 + 2xy - y^2) = \frac{{d}}{{dx}}(12)

Using the chain rule, the derivative of $x^2$ is $2x$ .

For the term $2xy$ , we must use the product rule. The derivative with respect to $x$ of $2xy$ is $2y + 2x \frac{{dy}}{{dx}}$ .

For the term $-y^2$ , we must use the chain rule. The derivative with respect to $x$ of $-y^2$ is $-2y \frac{{dy}}{{dx}}$ .

Thus, the left-hand side becomes:

2x + 2y + 2x \frac{{dy}}{{dx}} - 2y \frac{{dy}}{{dx}} = 0

Simplifying this equation:

2x + 2y + \frac{{d}}{{dx}}(2x \cdot y) - \frac{{d}}{{dx}}(y^2) = 0

2x + 2y + 2x \frac{{dy}}{{dx}} - 2y \frac{{dy}}{{dx}} = 0

2x \frac{{dy}}{{dx}} - 2y \frac{{dy}}{{dx}} = -2x - 2y

2\left(x - y\right) \frac{{dy}}{{dx}} = -2\left(x + y\right)

Finally, solving for $\frac{{dy}}{{dx}}$ :

\frac{{dy}}{{dx}} = \frac{{-2\left(x + y\right)}}{{2\left(x - y\right)}}

Simplifying:

\frac{{dy}}{{dx}} = -\frac{{x + y}}{{x - y}}

(b)

First, we find the slope by substituting $x = 4$ and $y = -2$ into $\frac{{dy}}{{dx}}$ :

m = \frac{{-4 + (-2)}}{{4 - (-2)}} = -\frac{{6}}{{6}} = -1

Now we have the slope $m = -1$ . To find the equation of the line, we use the point-slope form:

y - y_1 = m(x - x_1)

Substituting $x_1 = 4$ , $y_1 = -2$ , and $m = -1$ :

y - (-2) = -1(x - 4)

y + 2 = -x + 4

Simplifying:

y = -x + 2

Therefore, the equation of the tangent line to the curve at the point $(4, -2)$ is $y = -x + 2$ .

Answer: $\frac{{dy}}{{dx}} = -\frac{{x + y}}{{x - y}}$ and $y = -x + 2$