# Derivation of the Equation of an Ellipse (Grades 11-12)

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## Derivation of the Equation of an Ellipse

For the ellipse with foci [math]F_1 = (-c,0)[/math] and [math]F_2 = (c,0)[/math], derive the equation of the ellipse.

1.

Which of the following is the best definition of an ellipse?

- A squished circle.
- An oblong shaped closed curve that has two foci.
- Given two points, the foci, the set of all points where the sum of the the distances between a given point and the foci is constant.
- The set of all points that are bounded by the distances from two central points, foci, and the major and minor axes.

2.

Let [math](x,y)[/math] be a point on the ellipse. Which equation represents the distance, [math]d_1[/math], between [math]F_1[/math] and the point [math](x,y) ?[/math]

- [math]d_1 = sqrt( (x+c)^2 + y^2)[/math]
- [math]d_1 = sqrt( (x-c) + y^2)[/math]
- [math]d_1 = sqrt( (-x + c)^2 + y^2[/math]
- [math]d_1 = sqrt( (x-c)^2 + (y-x)^2)[/math]

3.

Which equation represents the distance, [math]d_2[/math], between [math]F_2[/math] and the point [math](x,y) ?[/math]

- [math]d_2 = sqrt( (c-x)^2 + (y-c)^2) [/math]
- [math]d_2 = sqrt( (x-c)^2 + y^2) [/math]
- [math]d_2 = sqrt( (c-x)^2)[/math]
- [math]d_2 = sqrt( (x+c)^2 + y^2)[/math]

4.

Let [math](-a,0)[/math] be the left vertex of the ellipse. What is the distance from this vertex to [math]F_1 ?[/math]

- [math]sqrt((c-a)^2 + y^2) [/math]
- [math]sqrt(a^2 + c^2)[/math]
- [math]a[/math]
- [math]a-c[/math]

5.

What is the distance between [math](-a,0)[/math] and [math]F_2 ?[/math]

- [math]a+c[/math]
- [math]c-a[/math]
- [math]a-c[/math]
- [math]sqrt(a^2 - c^2)[/math]

6.

Using the distances between the two foci and the vertex [math](-a,0)[/math], what does [math]d_1 + d_2[/math] equal?

- [math]d_1+ d_2 = 2a-c[/math]
- [math]d_1+d_2 = 2a-c[/math]
- [math]d_1+ d_2 = 2a[/math]
- [math]d_1 + d_2 = c[/math]

7.

Substituting the equations for [math]d_1, d_2[/math] found in the second and third questions into the above equation, and then moving the second square root term to the right side of the equation results in the following:

[math]sqrt( (x+c)^2 + y^2) = 2a - sqrt( (x-c)^2 + y^2)[/math]

The next step will be to square both sides. Why was the second square root term moved to the right side of the equation before squaring, instead of just squaring both sides immediately?

[math]sqrt( (x+c)^2 + y^2) = 2a - sqrt( (x-c)^2 + y^2)[/math]

The next step will be to square both sides. Why was the second square root term moved to the right side of the equation before squaring, instead of just squaring both sides immediately?

- It gets rid of all square root terms immediately.
- It will make the algebra easier later on, since the two square root terms are not multiplied together.
- This ensures that there won't be any multiple answers that usually result from square roots (taking the positive and negative).
- This makes that equation more balanced, which looks nicer.

8.

After squaring both sides, expanding, combing like terms, and dividing by any common constants, what is the resulting equation?

- [math]xc - a^2 = -a sqrt( (x-c)^2+y^2) [/math]
- [math] 2a^2 + c^2 + x^2 + y^2 = -2a sqrt( (x-c)^2+y^2) [/math]
- [math]-2a^2 + c^2 + x^2 + y^2 = 0[/math]
- [math]a^2 = cx[/math]

9.

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:

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?

[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?

- 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].
- 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].
- 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].
- 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.

10.

It can be shown that the value [math]b[/math] from the previous question relates to the vertices of the minor-axis. Specifically, the vertices are [math](0,-b)[/math] and [math](0,b)[/math]. Looking at the positive vertex, it forms an isosceles triangle with the two foci. What is the length of the two congruent sides? How is this related to the previous question?

- 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].

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