Math Problem Statement

find the equation of an ellipse with center at the origin and passing through points (1,2) and (3,1).

Solution

To find the equation of an ellipse with its center at the origin and passing through the points (1,2)(1,2) and (3,1)(3,1), we follow these steps:

Step 1: General Form of the Ellipse

The equation of an ellipse centered at the origin is given by: x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 where aa and bb are the semi-major and semi-minor axes, respectively.

Step 2: Substitute the Points into the Ellipse Equation

We know the ellipse passes through the points (1,2)(1,2) and (3,1)(3,1). Substituting these into the general equation gives us two equations:

  1. For the point (1,2)(1, 2): 12a2+22b2=1    1a2+4b2=1\frac{1^2}{a^2} + \frac{2^2}{b^2} = 1 \implies \frac{1}{a^2} + \frac{4}{b^2} = 1

  2. For the point (3,1)(3, 1): 32a2+12b2=1    9a2+1b2=1\frac{3^2}{a^2} + \frac{1^2}{b^2} = 1 \implies \frac{9}{a^2} + \frac{1}{b^2} = 1

Step 3: Solve the System of Equations

We now have the following system of equations: 1a2+4b2=1(Equation 1)\frac{1}{a^2} + \frac{4}{b^2} = 1 \quad \text{(Equation 1)} 9a2+1b2=1(Equation 2)\frac{9}{a^2} + \frac{1}{b^2} = 1 \quad \text{(Equation 2)}

To solve this system, let's multiply each equation by a2b2a^2b^2 to eliminate the denominators:

From Equation 1: b2+4a2=a2b2b^2 + 4a^2 = a^2b^2

From Equation 2: 9b2+a2=a2b29b^2 + a^2 = a^2b^2

Let's now solve these equations simultaneously to find a2a^2 and b2b^2.

Step 4: Eliminate a2b2a^2b^2 and Solve for a2a^2 and b2b^2

Subtract the first equation from the second: (9b2+a2)(b2+4a2)=0(9b^2 + a^2) - (b^2 + 4a^2) = 0 8b23a2=08b^2 - 3a^2 = 0 8b2=3a28b^2 = 3a^2 b2=38a2b^2 = \frac{3}{8}a^2

Step 5: Substitute Back to Find Specific Values

Substituting b2=38a2b^2 = \frac{3}{8}a^2 into Equation 1: 1a2+438a2=1\frac{1}{a^2} + \frac{4}{\frac{3}{8}a^2} = 1 1a2+323a2=1\frac{1}{a^2} + \frac{32}{3a^2} = 1 3+323a2=1\frac{3 + 32}{3a^2} = 1 353a2=1\frac{35}{3a^2} = 1 a2=353a^2 = \frac{35}{3}

Now, substitute back to find b2b^2: b2=38×353=358b^2 = \frac{3}{8} \times \frac{35}{3} = \frac{35}{8}

Step 6: Final Equation of the Ellipse

The equation of the ellipse is: x2353+y2358=1\frac{x^2}{\frac{35}{3}} + \frac{y^2}{\frac{35}{8}} = 1 This can be simplified as: 8x235+3y235=1\frac{8x^2}{35} + \frac{3y^2}{35} = 1 or multiplying the entire equation by 35: 8x2+3y2=358x^2 + 3y^2 = 35

This is the equation of the ellipse.

Would you like more details or have any questions? Here are five related questions you might find helpful:

  1. How do you derive the general form of an ellipse centered at a point other than the origin?
  2. What are the differences between ellipses and circles in terms of their equations?
  3. How do you find the foci of an ellipse?
  4. Can the major axis of an ellipse be aligned with the y-axis instead of the x-axis?
  5. How do the lengths of the axes affect the shape of the ellipse?

Tip: Remember that the general form of an ellipse is flexible and can be rotated or translated to fit different scenarios.

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Math Problem Analysis

Mathematical Concepts

Ellipses
Coordinate Geometry

Formulas

Equation of an ellipse

Theorems

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Suitable Grade Level

High School