Find the equation of the line tangent to the curve 2(x^2+y^2)^2 = 25(x^2-y^2) at the point (-3,1) when the equation of this tangent line is written in the form y=mx+b

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 \(y\).

Find the equation of the line tangent to the curve 2(x^2 + y^2)^2 = 25(x^2 - y^2) at the point (-3, 1) when the equation of this tangent line is written in the form y = mx + b.

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). This problem uses implicit differentiation and the point-slope form to derive the tangent line in the form y = mx + b. Suitable for students in Grades 11-12.

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