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To solve for the properties of the ellipse given by the equation \(16x^2 + 25y^2 - 64x - 336 = 0\), we need to rewrite the equation in standard form. Here are the steps to do this:

### 1. Rewriting the Equation
The given equation is:
\[ 16x^2 + 25y^2 - 64x - 336 = 0 \]

First, let's group the \(x\) and \(y\) terms:
\[ 16x^2 - 64x + 25y^2 - 336 = 0 \]

### 2. Completing the Square
#### For the \(x\)-terms:
\[ 16(x^2 - 4x) \]

To complete the square, we take \((4/2)^2 = 4\):
\[ 16(x^2 - 4x + 4 - 4) = 16((x-2)^2 - 4) = 16(x-2)^2 - 64 \]

#### For the \(y\)-terms:
\[ 25y^2 \]

Since there is no \(y\) linear term, it remains as it is.

So, the equation becomes:
\[ 16(x-2)^2 - 64 + 25y^2 - 336 = 0 \]

### 3. Simplifying the Equation
\[ 16(x-2)^2 + 25y^2 - 400 = 0 \]
\[ 16(x-2)^2 + 25y^2 = 400 \]
\[ \frac{16(x-2)^2}{400} + \frac{25y^2}{400} = 1 \]
\[ \frac{(x-2)^2}{25} + \frac{y^2}{16} = 1 \]

### 4. Standard Form of the Ellipse
The standard form of the ellipse is:
\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]

From the equation \(\frac{(x-2)^2}{25} + \frac{y^2}{16} = 1\), we can see that:
- \(h = 2\)
- \(k = 0\)
- \(a^2 = 25\), hence \(a = 5\)
- \(b^2 = 16\), hence \(b = 4\

Sketch the graph of an ellipse by determining its properties.

Learn how to transform the ellipse equation 16x^2 + 25y^2 - 64x - 336 = 0 into its standard form and determine its graph and properties, including the axes and foci. Suitable for students in Grades 10-12.

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