To solve this problem and determine the equation of the hyperbola in its canonical form, let's analyze the information step by step:

Step 1: General equation of a hyperbola

If the foci of the hyperbola lie symmetrically on the $x$-axis (horizontal hyperbola), the canonical equation of the hyperbola is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

• The distance from the center to each focus is $c$, where $c^2 = a^2 + b^2$.

Step 2: Given data

1. The foci lie symmetrically with respect to the origin on the $x$-axis, meaning the hyperbola's center is at the origin, and it opens along the $x$-axis.

2. Two points $M_1(4, -4\sqrt{3})$ and $M_2(3, 2\sqrt{5})$ lie on the hyperbola. This means these points satisfy the equation:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

We will use these points to determine $a^2$ and $b^2$.

Step 3: Substitution of points into the equation

1. Start with point $M_1(4, -4\sqrt{3})$: $\frac{4^2}{a^2} - \frac{(-4\sqrt{3})^2}{b^2} = 1$ Simplifying: $\frac{16}{a^2} - \frac{48}{b^2} = 1 \quad \text{(Equation 1)}$

2. Next, substitute point $M_2(3, 2\sqrt{5})$: $\frac{3^2}{a^2} - \frac{(2\sqrt{5})^2}{b^2} = 1$ Simplifying: $\frac{9}{a^2} - \frac{20}{b^2} = 1 \quad \text{(Equation 2)}$

Step 4: Solving the system of equations

We now solve the system of two equations:

1. $\frac{16}{a^2} - \frac{48}{b^2} = 1$
2. $\frac{9}{a^2} - \frac{20}{b^2} = 1$

Introduce substitutions:
Let $A = \frac{1}{a^2}$ and $B = \frac{1}{b^2}$. The equations become:

1. $16A - 48B = 1$
2. $9A - 20B = 1$

Solve for $A$ and $B$ using elimination or substitution.

• Multiply the second equation by 16 to align coefficients with the first equation:

  $16(9A - 20B) = 16 \quad \Rightarrow \quad 144A - 320B = 16$

• Subtract the first equation $(16A - 48B = 1)$ multiplied by 9:

  $9(16A - 48B) = 9 \quad \Rightarrow \quad 144A - 432B = 9$

Now subtract:

$(144A - 320B) - (144A - 432B) = 16 - 9$ $112B = 7 \quad \Rightarrow \quad B = \frac{7}{112} = \frac{1}{16}$

Thus:

$\frac{1}{b^2} = \frac{1}{16} \quad \Rightarrow \quad b^2 = 16$

Substitute $B = \frac{1}{16}$ into $16A - 48B = 1$:

$16A - 48\left(\frac{1}{16}\right) = 1$ $16A - 3 = 1 \quad \Rightarrow \quad 16A = 4 \quad \Rightarrow \quad A = \frac{4}{16} = \frac{1}{4}$

Thus:

$\frac{1}{a^2} = \frac{1}{4} \quad \Rightarrow \quad a^2 = 4$

Step 5: Final equation of the hyperbola

Now that we know $a^2 = 4$ and $b^2 = 16$, the equation of the hyperbola is:

$\frac{x^2}{4} - \frac{y^2}{16} = 1$

Final Answer:

The equation of the hyperbola is:

$\frac{x^2}{4} - \frac{y^2}{16} = 1$

Do you need further clarification or help with similar problems? Here are five related questions:

1. How do you derive the equation of a vertical hyperbola?
2. How are the foci and vertices of a hyperbola determined from its equation?
3. Solve for the equation of a hyperbola given its asymptotes and one point.
4. What is the general form of the equation of a hyperbola centered at $(h, k)$?
5. How do you calculate the eccentricity of a hyperbola?

Tip: Remember, for hyperbolas, $c^2 = a^2 + b^2$, where $c$ is the distance to the foci.