##### Question Info

This question is public and is used in 1 group and 12 tests or worksheets.

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

View all questions by nsharp1.

# Nonlinear Equations and Functions Question

View this question.

Add this question to a group or test by clicking the appropriate button below.

Note: This question is included in a group. The contents of the question may require the group's common instructions or reference text to be meaningful. If so, you may want to add the entire group of questions to your test. To do this, click on the group instructions in the blue box below. If you choose to add only this question, common instructions or reference text will not be added to your test.

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

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

 $a^2 x^2 - c^2 x^2 + a^2 y^2$ $=$ $a^4 -a^2 c^2$ $(a^2 - c^2)x^2 + a^2 y^2$ $=$ $(a^2 - c^2) a^2$ $b^2 x^2 + a^2 y^2$ $=$ $a^2 b^2$ $x^2 / a^2 + y^2 / b^2$ $=$ $1$

This is the formula for the equation of an ellipse centered at $(0,0)$. Why can the substitution $b^2 = a^2-c^2$ be made?
1. Since $b$ is undefined so far, it can be defined as any value. Then, using the Pythagorean theorem, $a^2 + b^2 = c^2$, simply rearrange to solve for $b^2$.
2. Knowing that the semi-minor axis is $b$ units long, one can substitute the square of this value for $a^2-c^2$.
3. Since $b$ is not yet defined, it can be used to simplify the equation by defining $b^2 = a^2 - c^2$. A positive value for $a^2-c^2$ exists since $|a| > |c|$.
4. Projecting $b$, the length of the semi-minor axis, onto the semi-major axis it is seen that $b=a-c$. Then, simply square both sides of the equation.
You need to have at least 5 reputation to vote a question down. Learn How To Earn Badges.