To find the equation of the tangent line to the curve $2(x^2 + y^2)^2 = 25(x^2 - y^2)$ at the point $(-3, 1)$, we’ll go through the following steps:

Step 1: Differentiate the Equation Implicitly

Given: $2(x^2 + y^2)^2 = 25(x^2 - y^2)$ Let’s implicitly differentiate both sides with respect to $x$. We will use the chain rule and the product rule as needed.

Differentiate the left side:

$\frac{d}{dx}\left(2(x^2 + y^2)^2\right) = 2 \cdot 2(x^2 + y^2) \cdot \frac{d}{dx}(x^2 + y^2)$ Now, differentiate $x^2 + y^2$ with respect to $x$: $\frac{d}{dx}(x^2 + y^2) = 2x + 2y \frac{dy}{dx}$ So, the derivative of the left side becomes: $4(x^2 + y^2)(x + y \frac{dy}{dx})$

Differentiate the right side:

$\frac{d}{dx}\left(25(x^2 - y^2)\right) = 25 \left(2x - 2y \frac{dy}{dx}\right) = 50x - 50y \frac{dy}{dx}$

Equate both derivatives:

$4(x^2 + y^2)(x + y \frac{dy}{dx}) = 50x - 50y \frac{dy}{dx}$

Step 2: Solve for $\frac{dy}{dx}$

Move all terms involving $\frac{dy}{dx}$ to one side: $4(x^2 + y^2) y \frac{dy}{dx} + 50y \frac{dy}{dx} = 50x - 4(x^2 + y^2)x$ Factor out $\frac{dy}{dx}$: $\frac{dy}{dx} \left(4y(x^2 + y^2) + 50y\right) = 50x - 4x(x^2 + y^2)$ Then, $\frac{dy}{dx} = \frac{50x - 4x(x^2 + y^2)}{4y(x^2 + y^2) + 50y}$

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

Now substitute $x = -3$ and $y = 1$ into the derivative:

1. Compute $x^2 + y^2$ at $(-3, 1)$: $x^2 + y^2 = (-3)^2 + (1)^2 = 9 + 1 = 10$

2. Substitute into the formula: $\frac{dy}{dx} = \frac{50(-3) - 4(-3)(10)}{4(1)(10) + 50(1)}$ Simplify each part:

   • The numerator: $50(-3) - 4(-3)(10) = -150 + 120 = -30$
   • The denominator: $4 \cdot 10 + 50 = 40 + 50 = 90$

   So, $\frac{dy}{dx} = \frac{-30}{90} = -\frac{1}{3}$

Therefore, the slope $m$ of the tangent line at the point $(-3, 1)$ is $-\frac{1}{3}$.

Step 4: Use the Point-Slope Form to Find the Equation of the Tangent Line

Now that we have the slope $m = -\frac{1}{3}$ and the point $(-3, 1)$, we can use the point-slope form of the line: $y - y_1 = m(x - x_1)$ Substitute $m = -\frac{1}{3}$, $x_1 = -3$, and $y_1 = 1$: $y - 1 = -\frac{1}{3}(x + 3)$ Expanding and simplifying: $y - 1 = -\frac{1}{3}x - 1$ $y = -\frac{1}{3}x$

Final Answer

The equation of the tangent line is: $y = -\frac{1}{3}x$

Do you need further explanation on any step, or have any additional questions?

Here are five related questions to expand your understanding:

1. How can we find the normal line to this curve at $(-3, 1)$?
2. How would the solution change if we wanted the tangent line at a different point on the curve?
3. What are some methods to check if $(-3, 1)$ actually lies on the given curve?
4. How can implicit differentiation help with curves that aren't easily expressed as $y = f(x)$?
5. How would this process differ if we used polar coordinates?

Tip: When implicitly differentiating, keep track of terms carefully—common mistakes often happen with the chain rule and differentiating products of functions involving $y$.