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\