To convert the given equation to its general form, we start with the given equation:

$\frac{(x - 3)^2}{16} + \frac{(y - 3)}{1} = 1$

The equation represents an ellipse with the horizontal and vertical axes scaled differently. Our goal is to express this equation in the general quadratic form:

$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$

Step 1: Clear the denominators

Multiply the entire equation by 16 to eliminate the fraction:

$(x - 3)^2 + 16(y - 3) = 16$

Step 2: Expand the terms

First, expand $(x - 3)^2$ and $16(y - 3)$:

$(x - 3)^2 = x^2 - 6x + 9$ $16(y - 3) = 16y - 48$

Thus, the equation becomes:

$x^2 - 6x + 9 + 16y - 48 = 16$

Step 3: Combine like terms

Now, simplify the constants:

$x^2 - 6x + 16y - 39 = 0$

Final Answer: General Form

The equation in general form is:

$x^2 - 6x + 16y - 39 = 0$

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Do you want more details or have any questions about this?

Here are 5 related questions:

1. How do you determine the center of an ellipse from its equation?
2. What are the axes of symmetry for this ellipse?
3. How would you convert this general form back to the standard form of an ellipse?
4. What is the significance of the coefficients in the general form of a conic section?
5. How do you find the vertices of an ellipse?

Tip: Always check for correct expansion of terms when converting between different forms of equations.