Find the Standard form, and the coordinates of the center, vertices, and the foci of 25x² + 169y² - 50x - 676y - 3,524 = 0

We are given the equation:

\[
25x^2 + 169y^2 - 50x - 676y - 3,524 = 0
\]

This is the general form of the equation of a conic section. To find the standard form, we need to complete the square for both \(x\) and \(y\).

### Step 1: Rearrange the equation
First, we group the \(x\) and \(y\) terms together and move the constant to the right side of the equation:

\[
25x^2 - 50x + 169y^2 - 676y = 3,524
\]

### Step 2: Complete the square
#### For the \(x\) terms:
Factor out the coefficient of \(x^2\):

\[
25(x^2 - 2x)
\]

Now, complete the square. Take half the coefficient of \(x\) (which is \(-2\)), square it, and add it inside the parentheses. Don't forget to balance the equation by subtracting the equivalent term outside:

\[
25\left(x^2 - 2x + 1 - 1\right) = 25\left((x - 1)^2 - 1\right) = 25(x - 1)^2 - 25
\]

#### For the \(y\) terms:
Factor out the coefficient of \(y^2\):

\[
169(y^2 - 4y)
\]

Complete the square for \(y\). Take half the coefficient of \(y\) (which is \(-4\)), square it, and add it inside the parentheses:

\[
169\left(y^2 - 4y + 4 - 4\right) = 169\left((y - 2)^2 - 4\right) = 169(y - 2)^2 - 676
\]

### Step 3: Write the completed equation
Now, substitute the completed squares back into the equation:

\[
25(x - 1)^2 - 25 + 169(y - 2)^2 - 676 = 3,524
\]

Simplify by combining the constants on the right side:

\[
25(x - 1)^2 + 169(y - 2)^2 - 701 = 3,524
\]

Add 701 to both sides:

\[
25(x - 1)^2 + 169(y - 2)^2 = 4,225
\]

### Step 4: Divide by 4,225 to get the standard form
To convert this into standard form, divide the entire equation by 4,225:

\[
\frac{25(x - 1)^2}{4,225} + \frac{169(y - 2)^2}{4,225} = 1
\]

Simplify the fractions:

\[
\frac{(x - 1)^2}{169} + \frac{(y - 2)^2}{25} = 1
\]

This is the standard form of the ellipse.

### Step 5: Find the center, vertices, and foci
The standard form of the ellipse is:

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

Comparing it with our equation:

- \(h = 1\), \(k = 2\)
- \(a^2 = 169\), so \(a = 13\)
- \(b^2 = 25\), so \(b = 5\)

The center of the ellipse is \((h, k) = (1, 2)\).

#### Vertices:
The vertices are \(a\) units from the center along the x-axis. So the vertices are:

\[
(1 \pm 13, 2) = (14, 2) \text{ and } (-12, 2)
\]

#### Foci:
The distance from the center to the foci is given by:

\[
c = \sqrt{a^2 - b^2} = \sqrt{169 - 25} = \sqrt{144} = 12
\]

So the foci are:

\[
(1 \pm 12, 2) = (13, 2) \text{ and } (-11, 2)
\]

### Final Answer:
- **Standard Form:** \(\frac{(x - 1)^2}{169} + \frac{(y - 2)^2}{25} = 1\)
- **Center:** \((1, 2)\)
- **Vertices:** \((14, 2)\) and \((-12, 2)\)
- **Foci:** \((13, 2)\) and \((-11, 2)\)

Would you like any more details on this solution or further clarifications?

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**5 Related Questions:**
1. How do you find the eccentricity of an ellipse?
2. What is the general form of a hyperbola, and how does it differ from that of an ellipse?
3. How can you find the equation of a tangent to an ellipse at a given point?
4. What are the real-world applications of ellipses?
5. How do you convert the standard form of an ellipse back to the general form?

**Tip:** When completing the square, always remember to balance the equation by adding or subtracting the equivalent value from both sides.

Discover how to find the standard form, center, vertices, and foci of the ellipse given by the equation 25x² + 169y² - 50x - 676y - 3,524 = 0. This detailed solution includes completing the square, properties of ellipses, and the formulas for foci. Suitable for Grades 10-12.

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