We are given the equation \( 9x^2 + y^2 + 18x - 4y - 23 = 0 \), and we can solve this by completing the square for both \(x\) and \(y\) to express it in a more familiar form.

### Step-by-step solution:

1. **Group the terms by variable:**

   \[
   (9x^2 + 18x) + (y^2 - 4y) = 23
   \]

2. **Complete the square for the \(x\)-terms:**

   Factor out the 9 from the \(x\)-terms:

   \[
   9(x^2 + 2x)
   \]

   Now, complete the square inside the parentheses:

   \[
   x^2 + 2x \quad \text{can be written as} \quad (x+1)^2 - 1
   \]

   So,

   \[
   9(x^2 + 2x) = 9((x+1)^2 - 1) = 9(x+1)^2 - 9
   \]

3. **Complete the square for the \(y\)-terms:**

   \[
   y^2 - 4y
   \]

   Completing the square:

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

4. **Substitute these completed square forms back into the equation:**

   \[
   9(x+1)^2 - 9 + (y-2)^2 - 4 = 23
   \]

   Combine constants:

   \[
   9(x+1)^2 + (y-2)^2 = 23 + 9 + 4 = 36
   \]

5. **Rewrite the equation in standard form:**

   Divide through by 36:

   \[
   \frac{9(x+1)^2}{36} + \frac{(y-2)^2}{36} = 1
   \]

   Simplify:

   \[
   \frac{(x+1)^2}{4} + \frac{(y-2)^2}{36} = 1
   \]

### Conclusion:
This is the equation of an **ellipse** in standard form:

\[
\frac{(x+1)^2}{4} + \frac{(y-2)^2}{36} = 1
\]

- The center of the ellipse is at \( (-1, 2) \).
- The semi-major axis length is 6 (along the y-axis).
- The semi-minor axis length is 2 (along the x-axis).

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

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Here are 5 related questions to explore:

1. How do you graph an ellipse given its equation?
2. How do you derive the formula for the foci of an ellipse?
3. What is the difference between ellipses, parabolas, and hyperbolas?
4. How do you solve systems of equations involving conic sections?
5. How do eccentricity and axes relate to the shape of an ellipse?

**Tip:** When completing the square, always remember to add the same constants to both sides to maintain equality.

Solve the conic section equation 9x^2 + y^2 + 18x - 4y - 23 = 0 by completing the square and transforming it into the standard form of an ellipse. This detailed step-by-step solution explains how to rewrite the equation and identify the center, axes, and shape of the ellipse. Suitable for students in Grades 10-12.

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