# 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

- They each have a length of [math]2c[/math]. Using the distance formula and looking at the difference of distances between the lengths just found and the other vertices of the major axes, one finds that [math]b = sqrt(a^2-c^2[/math].
- They each have a length of [math]c[/math]. Then, the right triangle formed between the vertices and the origin, and applying the Pythagorean theorem, results in [math]a^2 + b^2 = c^2[/math].
- They each have a length of [math]a[/math]. Looking at the right triangle formed by the origin, [math]F_1[/math], and the vertex [math](0,b)[/math], and applying the Pythagorean theorem results in [math]a^2 = b^2 + c^2[/math].
- They each have a length of [math]a/2[/math]. Therefore, the sum of their lengths, [math]a[/math], can be used as a value equal to the sum of the lengths of [math]b[/math] and [math]c[/math].