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can you give me the a^2 and b^2 of this formula

The given equation is:

\[
25x^2 - 4y^2 - 50x + 16y - 391 = 0
\]

This equation resembles the general form of a hyperbola:

\[
\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1
\]

We can rewrite the equation in standard form by completing the square for both \(x\) and \(y\).

### Step 1: Group the \(x\)-terms and \(y\)-terms
\[
25(x^2 - 2x) - 4(y^2 - 4y) = 391
\]

### Step 2: Complete the square
For the \(x\)-terms \(x^2 - 2x\):
- Take half of \(-2\) (which is \(-1\)) and square it. \((-1)^2 = 1\).
- Add and subtract 1 inside the parentheses:
\[
25((x - 1)^2 - 1)
\]
This becomes:
\[
25(x - 1)^2 - 25
\]

For the \(y\)-terms \(y^2 - 4y\):
- Take half of \(-4\) (which is \(-2\)) and square it. \((-2)^2 = 4\).
- Add and subtract 4 inside the parentheses:
\[
-4((y - 2)^2 - 4)
\]
This becomes:
\[
-4(y - 2)^2 + 16
\]

### Step 3: Substitute back into the equation
Now the equation becomes:
\[
25(x - 1)^2 - 25 - 4(y - 2)^2 + 16 = 391
\]

### Step 4: Simplify
\[
25(x - 1)^2 - 4(y - 2)^2 - 9 = 391
\]
Add 9 to both sides:
\[
25(x - 1)^2 - 4(y - 2)^2 = 400
\]

### Step 5: Divide by 400 to express in standard form
\[
\frac{(x - 1)^2}{16} - \frac{(y - 2)^2}{100} = 1
\]

Now, the equation is in standard form for a hyperbola, where:
- \(a^2 = 16\)
- \(b^2 = 100\)

Thus, \(a^2 = 16\) and \(b^2 = 100\).

Would you like a more detailed breakdown, or do you have any other questions? Here are some related questions to consider:

1. How do we determine the center of the hyperbola?
2. What is the orientation of this hyperbola (horizontal or vertical)?
3. How do we calculate the asymptotes of the hyperbola?
4. What are the vertices of the hyperbola based on the values of \(a\) and \(b\)?
5. How does completing the square help in rewriting conic sections into standard form?

**Tip**: When completing the square, remember to balance the equation by adding or subtracting constants on both sides.

can you give me the a^2 and b^2 of this formula?

Learn how to find the values of a^2 and b^2 for the hyperbola equation 25x^2 - 4y^2 - 50x + 16y - 391 by completing the square and rewriting it in standard form. Ideal for high school students in Grades 10-12.

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