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Type: Multiple-Choice
Category: Nonlinear Equations and Functions
Level: Grade 11
Standards: HSG-GPE.A.3
Author: nsharp1
Created: 3 years ago

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Nonlinear Equations and Functions Question

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Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.3

After some more algebra, and then moving all terms with [math]x[/math] or [math]y[/math] to the left side of the equation (and all others to the right), the resulting equation is:


[math] a^2 x^2 - c^2 x^2 + a^2 y^2 [/math][math] = [/math][math] a^4 -a^2 c^2[/math]
[math](a^2 - c^2)x^2 + a^2 y^2 [/math][math] = [/math][math] (a^2 - c^2) a^2 [/math]
[math] b^2 x^2 + a^2 y^2 [/math][math] = [/math][math] a^2 b^2 [/math]
[math]x^2 / a^2 + y^2 / b^2 [/math][math] = [/math][math] 1 [/math]


This is the formula for the equation of an ellipse centered at [math](0,0)[/math]. Why can the substitution [math]b^2 = a^2-c^2[/math] be made?
  1. Since [math]b[/math] is undefined so far, it can be defined as any value. Then, using the Pythagorean theorem, [math]a^2 + b^2 = c^2[/math], simply rearrange to solve for [math]b^2[/math].
  2. Knowing that the semi-minor axis is [math]b[/math] units long, one can substitute the square of this value for [math]a^2-c^2[/math].
  3. Since [math]b[/math] is not yet defined, it can be used to simplify the equation by defining [math]b^2 = a^2 - c^2[/math]. A positive value for [math]a^2-c^2[/math] exists since [math]|a| > |c|[/math].
  4. Projecting [math]b[/math], the length of the semi-minor axis, onto the semi-major axis it is seen that [math]b=a-c[/math]. Then, simply square both sides of the equation.
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