https://file.aiquickdraw.com/mathbot/file/1728802650302ced1kv7e.jpg

Let's work through the conic sections given in the problem and translate them so that their centers or vertices are at the origin.

### a) \((x - 3)^2 = 8(y - 2)\)
This equation represents a **parabola**. To translate the center to the origin, shift the vertex \((3, 2)\) to \((0, 0)\).

Translation:
- Let \( x' = x - 3 \) and \( y' = y - 2 \), so the equation becomes:
  \[
  (x')^2 = 8y'
  \]
This is now a parabola with vertex at \((0, 0)\), opening upwards.

---

### b) \(16(x + 3)^2 + 25y^2 = 400\)
This is an **ellipse**. To center it at the origin, shift the center \((-3, 0)\) to \((0, 0)\).

Translation:
- Let \( x' = x + 3 \) and \( y' = y \), so the equation becomes:
  \[
  16(x')^2 + 25y'^2 = 400
  \]
Divide by 400 to get it in standard form:
  \[
  \frac{(x')^2}{25} + \frac{y'^2}{16} = 1
  \]
This is now an ellipse centered at the origin with semi-major axis 5 along the \(x'\)-axis and semi-minor axis 4 along the \(y'\)-axis.

---

### c) \(y^2 - 4x - 2y + 9 = 0\)
This equation represents a **parabola**, but we need to rewrite it in a more recognizable form by completing the square.

1. First, rearrange the terms:
   \[
   y^2 - 2y = 4x - 9
   \]
2. Complete the square on \(y\):
   \[
   (y - 1)^2 = 4x - 8
   \]
3. Simplify:
   \[
   (y - 1)^2 = 4(x - 2)
   \]
Now, the vertex is at \((2, 1)\). To shift the vertex to the origin:
- Let \( x' = x - 2 \) and \( y' = y - 1 \), so the equation becomes:
  \[
  (y')^2 = 4x'
  \]
This is now a parabola with vertex at the origin, opening to the right.

---

### Summary of Translations:
- **a)** Translated to \((x')^2 = 8y'\), a parabola centered at the origin.
- **b)** Translated to \(\frac{(x')^2}{25} + \frac{y'^2}{16} = 1\), an ellipse centered at the origin.
- **c)** Translated to \((y')^2 = 4x'\), a parabola centered at the origin.

Would you like to see the graphs of these conics or need further details on the translation process?

Here are 5 related questions:
1. What are the general forms of equations for parabolas, ellipses, and hyperbolas?
2. How do you identify the vertex or center of a conic section from its equation?
3. How does completing the square help in rewriting conic equations?
4. What are the key characteristics of ellipses and parabolas (axes, direction of opening, etc.)?
5. How can you determine whether a conic is a parabola, ellipse, or hyperbola from its equation?

**Tip:** For parabolas, the coefficient of the squared term indicates whether the parabola opens horizontally or vertically.

Translate so that the center or vertex of the conic section is at the origin of the new coordinate system. Sketch the curve showing both the old and new axes for the following: 
a) (x - 3)^2 = 8(y - 2) 
b) 16(x + 3)^2 + 25y^2 = 400 
c) y^2 - 4x - 2y + 9 = 0

Learn how to translate conic sections such as parabolas and ellipses to center them at the origin. This problem covers translating the axes and completing the square for equations like (x - 3)^2 = 8(y - 2) and 16(x + 3)^2 + 25y^2 = 400. Suitable for students in Grades 10-12.

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