Let's go through the steps required to solve the problem:

Part a: Determine the equation of the tangent line and the normal line at the given point (6, 5) on the curve.

The given equation is: $\left( x^2 + y^2 \right)^2 = 3721 \left( x^2 - y^2 \right)$

We'll follow these steps:

1. Implicit Differentiation: Differentiate both sides of the equation with respect to $x$ to find the slope of the tangent line.
2. Find the Slope of the Tangent Line: Use the point (6, 5) to calculate the slope.
3. Equation of the Tangent Line: Use the point-slope form of the line equation.
4. Equation of the Normal Line: The normal line is perpendicular to the tangent line. If the slope of the tangent line is $m$, the slope of the normal line is $-\frac{1}{m}$.

Step 1: Implicit Differentiation

Differentiate both sides of the equation with respect to $x$.

The left-hand side is: $\frac{d}{dx} \left( x^2 + y^2 \right)^2 = 2 \left( x^2 + y^2 \right) \cdot \left( 2x + 2y \frac{dy}{dx} \right)$ The right-hand side is: $\frac{d}{dx} \left( 3721 \left( x^2 - y^2 \right) \right) = 3721 \left( 2x - 2y \frac{dy}{dx} \right)$

This results in: $2 \left( x^2 + y^2 \right) \cdot \left( 2x + 2y \frac{dy}{dx} \right) = 3721 \left( 2x - 2y \frac{dy}{dx} \right)$

Simplify: $4 \left( x^2 + y^2 \right) \cdot \left( x + y \frac{dy}{dx} \right) = 7442 \left( x - y \frac{dy}{dx} \right)$

Step 2: Find the Slope at $(6, 5)$

Now, substitute $x = 6$ and $y = 5$ into the differentiated equation. This will allow us to find $\frac{dy}{dx}$, which is the slope of the tangent line at the point (6, 5).

Substitute $x = 6$ and $y = 5$ into the original equation to find the values of $x^2 + y^2$: $(6^2 + 5^2)^2 = 3721 \cdot (6^2 - 5^2)$ Simplifying this will allow us to calculate the slope of the tangent line.

Let me compute this for you.The slope of the tangent line at the point $(6, 5)$ is $\frac{118}{105}$.

Step 3: Equation of the Tangent Line

We can now use the point-slope form of the equation of a line:

$y - y_1 = m (x - x_1)$

Substituting $m = \frac{118}{105}$ and the point $(6, 5)$:

$y - 5 = \frac{118}{105}(x - 6)$

This is the equation of the tangent line.

Step 4: Equation of the Normal Line

The slope of the normal line is the negative reciprocal of the tangent slope:

$m_{\text{normal}} = -\frac{1}{\frac{118}{105}} = -\frac{105}{118}$

Using the point-slope form again for the normal line:

$y - 5 = -\frac{105}{118}(x - 6)$

This is the equation of the normal line.

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Part b: Graph the Tangent and Normal Lines

Would you like me to plot the graph of the curve along with the tangent and normal lines at the point $(6, 5)$?