The Ellipse

Definition of an Ellipse

An ellipse is the locus of a point P that moves in a plane such that the sum of its distances from two fixed points (the foci, F₁ and F₂) is constant. This constant sum is equal to the length of the major axis (2a).

$$PF_1 + PF_2 = 2a$$

Alternatively, it can be defined as a conic section with an eccentricity e such that $0 < e < 1$. An eccentricity of 0 is a perfect circle, and as 'e' approaches 1, the ellipse becomes more elongated.

Standard Equation of an Ellipse

The standard equation assumes the ellipse is centered at the origin (0,0).

• Horizontal Ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b$)

  • Vertices: $(\pm a, 0)$

  • Foci: $(\pm c, 0)$, where $c=ae$.

  • Major Axis: Length $2a$ (along the x-axis).

  • Minor Axis: Length $2b$ (along the y-axis).

• Vertical Ellipse: $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ (where $a > b$)

  • Vertices: $(0, \pm a)$

  • Foci: $(0, \pm c)$, where $c=ae$.

  • Major Axis: Length $2a$ (along the y-axis).

  • Minor Axis: Length $2b$ (along the x-axis).

Key Terminology and Formulas

For a horizontal ellipse (similar formulas apply for a vertical one):

• Relationship between a, b, and c: The distances from the center to a vertex (a), center to a co-vertex (b), and center to a focus (c) are related by a formula resembling the Pythagorean theorem:

$$c^2 = a^2 - b^2$$

This also gives the key eccentricity relationship: 

$$b^2 = a^2(1-e^2)$$

• Directrices: The equations of the two directrices are $x = \pm \frac{a}{e}$.

• Latus Rectum: The focal chord perpendicular to the major axis. Its length is a measure of the ellipse's 'width' at the focus.

  • Length of the latus rectum: $\frac{2b^2}{a}$

  • Endpoints of the latus rectum: $(\pm ae, \pm \frac{b^2}{a})$

Parametric Coordinates

Any point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ can be represented by the coordinates $(a\cos\theta, b\sin\theta)$, where '$\theta$' is the eccentric angle, not the actual angle to the point.

Example 1: Finding Parameters from the Equation

Question: For the ellipse $16x^2+25y^2=400$, find the lengths of the major and minor axes, the coordinates of the foci, and the eccentricity.

Solution:

Step 1: Convert to standard form by dividing by 400:

$$\frac{16x^2}{400} + \frac{25y^2}{400} = 1 \implies \frac{x^2}{25} + \frac{y^2}{16} = 1$$

Step 2: Identify a and b. Since $25 > 16$, this is a horizontal ellipse with $a^2=25 \implies a=5$ and $b^2=16 \implies b=4$.

• Major Axis: $2a = \mathbf{10}$.

• Minor Axis: $2b = \mathbf{8}$.

Step 3: Find c and the foci.

$c^2 = a^2-b^2 = 25-16=9 \implies c=3$. The foci are at $(\pm c, 0) = \mathbf{(\pm 3, 0)}$.

Step 4: Find the eccentricity.

$e = c/a = \mathbf{3/5}$.

Example 2: Finding the Equation from Parameters

Question: Find the equation of the ellipse whose foci are at $(\pm 5, 0)$ and vertices are at $(\pm 13, 0)$.

Solution:

From the vertices $(\pm a, 0)$, we get $\mathbf{a=13}$. From the foci $(\pm c, 0)$, we get $\mathbf{c=5}$. Since the foci are on the x-axis, it's a horizontal ellipse. We need to find $b^2$.

Using the relation $c^2=a^2-b^2$:

$5^2 = 13^2 - b^2 \implies 25 = 169 - b^2 \implies b^2 = 169 - 25 = \mathbf{144}$.

The equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{x^2}{169} + \frac{y^2}{144} = 1$.

Example 3: Using the Latus Rectum

Question: Find the equation of the ellipse whose latus rectum is 10 and whose minor axis is equal to the distance between its foci.

Solution:

Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Condition 1: Length of latus rectum: $\frac{2b^2}{a} = 10 \implies b^2=5a$. (i)

Condition 2: Length of minor axis = Distance between foci: $2b = 2c \implies b=c$. (ii)

Now we use the main relation $c^2=a^2-b^2$. Substituting $c=b$ gives $b^2=a^2-b^2 \implies a^2=2b^2$. (iii)

Now we have a system of equations. Substitute (i) into (iii):

$a^2 = 2(5a) = 10a$. Since $a \n e 0$, we can divide by a to get $\mathbf{a=10}$.

Now find $b^2$ using (i): $b^2 = 5(10) = \mathbf{50}$.

The equation is $\frac{x^2}{100} + \frac{y^2}{50} = 1$.

Example 4: Ellipse with a Shifted Center

Question: Find the center, foci, and eccentricity of the ellipse $4x^2 + 9y^2 - 16x + 18y - 11 = 0$.

Solution:

Step 1: Group terms and complete the square.

$4(x^2 - 4x) + 9(y^2 + 2y) = 11$

$4(x^2 - 4x + 4) + 9(y^2 + 2y + 1) = 11 + 4(4) + 9(1)$

$4(x-2)^2 + 9(y+1)^2 = 36$

Step 2: Divide by 36 to get standard form.

$\frac{(x-2)^2}{9} + \frac{(y+1)^2}{4} = 1$

• Center: $(h,k) = \mathbf{(2, -1)}$.

• Parameters: $a^2=9 \implies a=3$; $b^2=4 \implies b=2$.

• Foci distance (c): $c^2 = a^2 - b^2 = 9 - 4 = 5 \implies c=\sqrt{5}$.

• Foci: The ellipse is horizontal, so foci are at $(h \pm c, k) = \mathbf{(2 \pm \sqrt{5}, -1)}$.

• Eccentricity (e): $e = c/a = \mathbf{\frac{\sqrt{5}}{3}}$.

Example 5: Using the Fundamental Definition

Question: A point P moves such that the sum of its distances from (4,0) and (-4,0) is 10. Find the equation of the locus of P.

Solution:

This is the definition of an ellipse. The two fixed points are the foci, and the constant sum is the length of the major axis.

• Foci: $(\pm 4, 0)$, so $\mathbf{c=4}$.

• Major Axis: $2a = 10$, so $\mathbf{a=5}$.

We find $b^2$ using $c^2=a^2-b^2$: $4^2=5^2-b^2 \implies 16=25-b^2 \implies b^2=9$.

The locus is a horizontal ellipse centered at the origin with the equation $\frac{x^2}{25} + \frac{y^2}{9} = 1$.