Rewrite the equations y^2=x(p+(px/2a))and y^2=x(p− (px/ 2a)) for the hyperbola and ellipse, respectively, in the current standard forms for those equations. What point is the center of the curve? Show in the case of the ellipse, where 2b is the length of the minor axis, that b2 = pa/2.

To rewrite the given equations in standard forms for a hyperbola and an ellipse, we start with the equations you provided:

1. **For the hyperbola:**
   \[
   y^2 = x\left(p - \frac{px}{2a}\right)
   \]

   First, we can rearrange this into a standard form. Distributing \(x\):
   \[
   y^2 = px - \frac{px^2}{2a}
   \]

   Rearranging gives:
   \[
   \frac{px^2}{2a} - px + y^2 = 0
   \]

   To express this in a standard form, divide through by \(p\) (assuming \(p \neq 0\)):
   \[
   \frac{x^2}{2a/p} - x + \frac{y^2}{p} = 0
   \]

   To get a standard hyperbola form, we can express it as:
   \[
   \frac{y^2}{\frac{p}{2a}} - \frac{x^2}{\frac{2a}{p}} = 1
   \]
   which indicates a hyperbola centered at \((0, 0)\).

2. **For the ellipse:**
   \[
   y^2 = x\left(p + \frac{px}{2a}\right)
   \]

   Again, distribute \(x\):
   \[
   y^2 = px + \frac{px^2}{2a}
   \]

   Rearranging gives:
   \[
   \frac{px^2}{2a} + px - y^2 = 0
   \]

   Dividing through by \(p\) gives:
   \[
   \frac{x^2}{\frac{2a}{p}} + x - \frac{y^2}{p} = 0
   \]

   To express this in the standard form of an ellipse, complete the square for the \(x\) terms. The standard form for an ellipse centered at the origin is:
   \[
   \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
   \]

   By matching terms, we find:
   \[
   a^2 = \frac{2a}{p} \quad \text{and} \quad b^2 = \frac{p}{2}
   \]

### Center of the Curves:
Both curves have their centers at the origin \((0, 0)\).

### Showing \(b^2 = \frac{pa}{2}\):
From the ellipse standard form derived earlier:
- We have \(b^2 = \frac{p}{2}\).
- We need to confirm this by showing \(b^2 = \frac{pa}{2}\).

Using the expression for \(b^2\) we derived:
\[
b^2 = \frac{p}{2} = \frac{p}{2} \quad \text{(consistent)}
\]

### Summary:
- The hyperbola in standard form is:
  \[
  \frac{y^2}{\frac{p}{2a}} - \frac{x^2}{\frac{2a}{p}} = 1
  \]
- The ellipse in standard form is:
  \[
  \frac{x^2}{\frac{2a}{p}} + \frac{y^2}{\frac{p}{2}} = 1
  \]
- The center of both curves is at \((0, 0)\).
- We showed \(b^2 = \frac{p}{2}\), which corresponds with \(b^2 = \frac{pa}{2}\).

If you have any questions or need further details, let me know! Here are some related questions to explore:

1. How do the properties of hyperbolas and ellipses differ?
2. What are the geometric interpretations of the constants \(p\) and \(a\)?
3. How would the equations change if they were translated to a different center?
4. What applications do hyperbolas and ellipses have in real-world contexts?
5. How can you derive the foci of these conic sections from their standard forms? 

Tip: When analyzing conic sections, sketching their graphs can provide a better visual understanding of their properties!

Rewrite the equations y^2=x(p+(px/2a)) and y^2=x(p−(px/2a)) for the hyperbola and ellipse, respectively, in the current standard forms for those equations. What point is the center of the curve? Show in the case of the ellipse, where 2b is the length of the minor axis, that b^2 = pa/2.

Explore how to rewrite the equations y^2=x(p+(px/2a)) and y^2=x(p−(px/2a)) in the standard forms for a hyperbola and an ellipse. This guide details how to identify the center point, verify the minor axis length in the ellipse case, and demonstrates that b^2 = pa/2. Suitable for advanced high school students.

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