Math Problem Statement

Find an equation for the hyperbola described. Graph the equation by hand.

Center at​ (0,0); focus at

​(00​,1010​);

vertex at

​(00​,negative 6−6​)

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Part 1

An equation of the hyperbola is

enter your response hereminus−enter your response hereequals=1

​(Use integers or fractions for any numbers in the​ equation.)

Find an equation for the ellipse. Graph the equation.

center at​ (0,0); focus at

​(0,2424​);

vertex at

​(0,negative 25−25​)

Question content area bottom

Part 1

Type the left side of the equation of the ellipse.

enter your response hereequals=1

Solution

Problem 1: Equation of the Hyperbola

Given:

  • Center: (0,0)(0, 0)
  • Focus: (0,±10)(0, \pm 10)
  • Vertex: (0,±6)(0, \pm 6)

Step 1: Hyperbola Orientation and Standard Form

The hyperbola opens vertically because both the focus and vertex are along the yy-axis. The standard form of such a hyperbola is:
y2a2x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1
Here:

  • aa is the distance from the center to a vertex.
  • cc is the distance from the center to a focus.
  • The relationship c2=a2+b2c^2 = a^2 + b^2 holds for hyperbolas.

Step 2: Determine a2a^2, c2c^2, and b2b^2

  • a=6a = 6, so a2=36a^2 = 36.
  • c=10c = 10, so c2=100c^2 = 100.
    Using c2=a2+b2c^2 = a^2 + b^2:
    100=36+b2    b2=64.100 = 36 + b^2 \implies b^2 = 64.

Step 3: Equation of the Hyperbola

Substitute a2a^2 and b2b^2 into the standard form:
y236x264=1.\frac{y^2}{36} - \frac{x^2}{64} = 1.

Problem 2: Equation of the Ellipse

Given:

  • Center: (0,0)(0, 0)
  • Focus: (0,±24)(0, \pm 24)
  • Vertex: (0,±25)(0, \pm 25)

Step 1: Ellipse Orientation and Standard Form

The ellipse is vertical, so the standard form is:
y2a2+x2b2=1\frac{y^2}{a^2} + \frac{x^2}{b^2} = 1
Here:

  • aa is the distance from the center to a vertex.
  • cc is the distance from the center to a focus.
  • The relationship c2=a2b2c^2 = a^2 - b^2 holds for ellipses.

Step 2: Determine a2a^2, c2c^2, and b2b^2

  • a=25a = 25, so a2=625a^2 = 625.
  • c=24c = 24, so c2=576c^2 = 576.
    Using c2=a2b2c^2 = a^2 - b^2:
    576=625b2    b2=49.576 = 625 - b^2 \implies b^2 = 49.

Step 3: Equation of the Ellipse

Substitute a2a^2 and b2b^2 into the standard form:
y2625+x249=1.\frac{y^2}{625} + \frac{x^2}{49} = 1.


Final Equations

  1. Hyperbola:
    y236x264=1.\frac{y^2}{36} - \frac{x^2}{64} = 1.

  2. Ellipse:
    y2625+x249=1.\frac{y^2}{625} + \frac{x^2}{49} = 1.

Would you like further assistance with graphing these equations?


Related Questions:

  1. What is the geometric significance of aa, bb, and cc in hyperbolas and ellipses?
  2. How do you differentiate between the equations of ellipses and hyperbolas?
  3. What steps are involved in graphing a hyperbola by hand?
  4. How do the foci influence the shape of an ellipse or hyperbola?
  5. What are the asymptotes of the hyperbola, and how are they calculated?

Tip: For hyperbolas, always determine the slopes of the asymptotes—they guide accurate graphing!

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Math Problem Analysis

Mathematical Concepts

Conic Sections
Hyperbolas
Ellipses
Standard Forms of Equations

Formulas

Hyperbola equation: (y^2/a^2) - (x^2/b^2) = 1
Ellipse equation: (y^2/a^2) + (x^2/b^2) = 1
Relationship for hyperbolas: c^2 = a^2 + b^2
Relationship for ellipses: c^2 = a^2 - b^2

Theorems

Properties of Conic Sections
Distance Formula

Suitable Grade Level

Grades 10-12