Math Problem Statement

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.

Solution

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: y2=x(ppx2a)y^2 = x\left(p - \frac{px}{2a}\right)

    First, we can rearrange this into a standard form. Distributing xx: y2=pxpx22ay^2 = px - \frac{px^2}{2a}

    Rearranging gives: px22apx+y2=0\frac{px^2}{2a} - px + y^2 = 0

    To express this in a standard form, divide through by pp (assuming p0p \neq 0): x22a/px+y2p=0\frac{x^2}{2a/p} - x + \frac{y^2}{p} = 0

    To get a standard hyperbola form, we can express it as: y2p2ax22ap=1\frac{y^2}{\frac{p}{2a}} - \frac{x^2}{\frac{2a}{p}} = 1 which indicates a hyperbola centered at (0,0)(0, 0).

  2. For the ellipse: y2=x(p+px2a)y^2 = x\left(p + \frac{px}{2a}\right)

    Again, distribute xx: y2=px+px22ay^2 = px + \frac{px^2}{2a}

    Rearranging gives: px22a+pxy2=0\frac{px^2}{2a} + px - y^2 = 0

    Dividing through by pp gives: x22ap+xy2p=0\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 xx terms. The standard form for an ellipse centered at the origin is: x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

    By matching terms, we find: a2=2apandb2=p2a^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)(0, 0).

Showing b2=pa2b^2 = \frac{pa}{2}:

From the ellipse standard form derived earlier:

  • We have b2=p2b^2 = \frac{p}{2}.
  • We need to confirm this by showing b2=pa2b^2 = \frac{pa}{2}.

Using the expression for b2b^2 we derived: b2=p2=p2(consistent)b^2 = \frac{p}{2} = \frac{p}{2} \quad \text{(consistent)}

Summary:

  • The hyperbola in standard form is: y2p2ax22ap=1\frac{y^2}{\frac{p}{2a}} - \frac{x^2}{\frac{2a}{p}} = 1
  • The ellipse in standard form is: x22ap+y2p2=1\frac{x^2}{\frac{2a}{p}} + \frac{y^2}{\frac{p}{2}} = 1
  • The center of both curves is at (0,0)(0, 0).
  • We showed b2=p2b^2 = \frac{p}{2}, which corresponds with b2=pa2b^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 pp and aa?
  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!

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Math Problem Analysis

Mathematical Concepts

Conic Sections
Standard Form of Hyperbola
Standard Form of Ellipse
Algebraic Manipulation

Formulas

Standard form of hyperbola: (y^2 / b^2) - (x^2 / a^2) = 1
Standard form of ellipse: (x^2 / a^2) + (y^2 / b^2) = 1
Minor axis length for ellipse: 2b
Relationship for ellipse: b^2 = pa / 2

Theorems

Standard Forms of Conic Sections

Suitable Grade Level

Grades 11-12