The Ellipse

Like the graph of a parabola, the graph of an ellipse can be translated. If an ellipse centered at $(0,0)$ is translated $h$ units horizontally and $k$ units vertically, then the center of the ellipse will be $(h, k)$. This translation changes the standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ by replacing $x$ with $(x - h)$ and $y$ with $(y - k)$.

Thus, the standard form of the equation of an ellipse with center $(h, k)$ is:

$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \quad \text{(Equation 1.12)}$

The standard equation of the translated ellipse $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$, for $a < b$ has the following features:

1. The vertices are $(h, k \pm b)$ and co-vertices are $(h \pm a, k)$.
2. The foci are $(h, k \pm be)$.
3. The directrices are $y = k \pm \frac{b}{e}$.
4. The lengths of major and minor axes are $2b$ and $2a$, respectively.
5. The eccentricity, $e = \frac{\sqrt{b^2 - a^2}}{b}$.

Example 1

Given the ellipse $\frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{4} = 1$. Find each of the following:

1. Center
2. Foci
3. Eccentricity
4. Directrices
5. Vertices
6. Length of major axis

Solution:

(a) Comparing with the standard equation $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$, we have $a = 3$, $b = 2$, $h = 2$, and $k = -1$. Therefore, the center is at $(2, -1)$.

(b) Since $a > b$, $c^2 = a^2 - b^2 = 9 - 4 = 5$, so $c = \sqrt{5}$. The foci are $(h \pm c, k) = (2 \pm \sqrt{5}, -1)$.

(c) $e = \frac{c}{a} = \frac{\sqrt{5}}{3}$

(d) The directrices are $x = h \pm \frac{a}{e} = 2 \pm \frac{3}{\frac{\sqrt{5}}{3}} = 2 \pm \frac{9}{\sqrt{5}} = 2 \pm \frac{9\sqrt{5}}{5}$

(e) The vertices are $(h \pm a, k) = (2 \pm 3, -1)$, which are $(5, -1)$ and $(-1, -1)$.

(f) Length of major axis $2a = 2(3) = 6$.

Example 2

Given the ellipse $25x^2 + 9y^2 - 100x - 18y - 116 = 0$. Find each of the following:

(a) Center (b) Foci (c) Vertices (d) Directrices

In order to sketch the graph, identify the x and y-intercepts as follows:

When x = 0, $9y^2 - 18y - 116 = 0$. This is a quadratic equation in y. Using the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{18 \pm \sqrt{(-18)^2 - 4(9)(-116)}}{2(9)} = \frac{18 \pm \sqrt{324 + 4176}}{18} = \frac{18 \pm \sqrt{4500}}{18} = \frac{18 \pm 30\sqrt{5}}{18} = 1 \pm \frac{5\sqrt{5}}{3}$

So, the y-intercepts are approximately $(0, -2.7)$ and $(0, 4.7)$.

When y = 0, $25x^2 - 100x - 116 = 0$. This is a quadratic equation in x. Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{100 \pm \sqrt{(-100)^2 - 4(25)(-116)}}{2(25)} = \frac{100 \pm \sqrt{10000 + 11600}}{50} = \frac{100 \pm \sqrt{21600}}{50} = \frac{100 \pm 60\sqrt{6}}{50} = 2 \pm \frac{6\sqrt{6}}{5}$

So, the x-intercepts are approximately $(-0.9, 0)$ and $(4.9, 0)$.

Solution (Calculations):

(a) Rearrange and complete the square:

$25(x^2 - 4x) + 9(y^2 - 2y) = 116$

$25(x^2 - 4x + 4) + 9(y^2 - 2y + 1) = 116 + 25(4) + 9(1)$

$25(x - 2)^2 + 9(y - 1)^2 = 116 + 100 + 9$

$25(x - 2)^2 + 9(y - 1)^2 = 225$

Divide by 225: $\frac{(x - 2)^2}{9} + \frac{(y - 1)^2}{25} = 1$

Comparing with the translated equation, $a = 3$, $b = 5$, $h = 2$, $k = 1$. Thus, the center is $(2, 1)$.

(b) Since $b > a$, $c^2 = b^2 - a^2 = 25 - 9 = 16$, so $c = 4$. The foci are $(h, k \pm c) = (2, 1 \pm 4)$, which are $(2, 5)$ and $(2, -3)$.

(c) The vertices are $(h, k \pm b) = (2, 1 \pm 5)$, which are $(2, 6)$ and $(2, -4)$.

(d) $e = \frac{c}{b} = \frac{4}{5}$. The directrices are $y = k \pm \frac{b}{e} = 1 \pm \frac{5}{\frac{4}{5}} = 1 \pm \frac{25}{4}$, which are $y = \frac{29}{4}$ and $y = -\frac{21}{4}$.

Similar to the parabola, we can determine the equations of the tangent and normal to an ellipse at any given point. The gradient of a tangent at the point of tangency is the same as the gradient of the ellipse at that point.

One way to find the equation of the tangent and normal is by differentiating the equation of the ellipse to determine its gradient at the point of tangency.

The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $P(x_1, y_1)$ is:

$\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$

The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $P(x_1, y_1)$ is:

$\frac{(y - y_1)b^2}{y_1} = \frac{(x - x_1)a^2}{x_1} \quad \text{or} \quad \frac{x-x_1}{x_1/a^2} = \frac{y-y_1}{y_1/b^2}$

Example 1

Find the equations of the tangent and normal to the ellipse $x^2 + 4y^2 = 100$ at the point $(6, 4)$.

Solution:

Standard form: $\frac{x^2}{100} + \frac{y^2}{25} = 1$. Here $a = 10$ and $b = 5$. The point of tangency is $(x_1, y_1) = (6, 4)$.

Tangent: $\frac{6x}{100} + \frac{4y}{25} = 1 \Rightarrow \frac{3x}{50} + \frac{4y}{25} = 1 \Rightarrow 3x + 8y = 50$

Normal: $\frac{(y - 4)25}{4} = \frac{(x - 6)100}{6} \Rightarrow 150(y-4)=400(x-6) \Rightarrow 3(y-4)=8(x-6)\Rightarrow 3y-12=8x-48 \Rightarrow 8x-3y-36=0$

Example 2

Find the equations of the tangent and normal to the ellipse $25x^2 + 9y^2 - 100x - 18y - 116 = 0$ at the point $(5,1)$.

Solution:

Rewrite the equation: $25(x-2)^2+9(y-1)^2=225$ or $\frac{(x-2)^2}{9} + \frac{(y-1)^2}{25} = 1$

Differentiating implicitly with respect to x:

$\frac{50(x-2)}{9} + \frac{18(y-1)}{25} \frac{dy}{dx} = 0$

At $(5, 1)$: $\frac{50(3)}{9} + 0 = 0$ which is incorrect. This means the tangent is vertical. The point is on the ellipse so the tangent is x=5 and the normal is y=1.

Using the translated formula: $\frac{(x-2)(5-2)}{9} + \frac{(y-1)(1-1)}{25} = 1 \Rightarrow \frac{3(x-2)}{9}=1 \Rightarrow x-2=3 \Rightarrow x=5$

Example 3

Prove that the condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$.

Solution:

Substitute $y = mx + c$ into $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

$\frac{x^2}{a^2} + \frac{(mx + c)^2}{b^2} = 1$

$b^2x^2 + a^2(m^2x^2 + 2mcx + c^2) = a^2b^2$

$(b^2 + a^2m^2)x^2 + 2a^2mcx + a^2c^2 - a^2b^2 = 0$

For tangency, the discriminant must be zero:

$(2a^2mc)^2 - 4(b^2 + a^2m^2)(a^2c^2 - a^2b^2) = 0$

$4a^4m^2c^2 - 4(a^2b^2c^2 - a^2b^4 + a^4m^2c^2 - a^4m^2b^2) = 0$

$-4a^2b^2c^2 + 4a^2b^4 + 4a^4m^2b^2 = 0$

Divide by $-4a^2b^2$:

$c^2 - b^2 - a^2m^2 = 0$

$c^2 = a^2m^2 + b^2$

The line segment $LL'$ is the latus rectum of the ellipse.

Since the coordinate point of the focus is $F(ae, 0)$, the two coordinate points $L(ae, y)$ and $L'(ae, -y)$ satisfy the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Substituting $x = ae$ into the equation of the ellipse gives:

$\frac{(ae)^2}{a^2} + \frac{y^2}{b^2} = 1$

$e^2 + \frac{y^2}{b^2} = 1$

Solving for $y$:

$\frac{y^2}{b^2} = 1 - e^2$

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

$y = \pm b\sqrt{1 - e^2}$

Thus, the coordinates of $L$ and $L'$ are $(ae, b\sqrt{1 - e^2})$ and $(ae, -b\sqrt{1 - e^2})$, respectively.

Length of latus rectum = length of $LL'$:

$LL' = b\sqrt{1 - e^2} - (-b\sqrt{1 - e^2}) = 2b\sqrt{1 - e^2}$

But, $b^2 = a^2(1 - e^2)$ which means $\sqrt{1-e^2} = \frac{b}{a}$.

Substitute this result into the equation of the length of the latus rectum:

$LL' = 2b\left(\frac{b}{a}\right) = \frac{2b^2}{a}$

Therefore, the length of the latus rectum is $\frac{2b^2}{a}$ units for $a > b$. A similar approach can be used to show that the length of the latus rectum is $\frac{2a^2}{b}$ units for $b > a$.

Example 1

The lengths of the latus rectum and major axis of the ellipse are $\frac{64}{5}$ units and 20 units, respectively. If the major axis is along the y-axis, find the equation of the ellipse.

Solution:

Since the major axis is along the y-axis, the length of the latus rectum is $\frac{2a^2}{b}$. We are given that this length is $\frac{64}{5}$, so:

$\frac{2a^2}{b} = \frac{64}{5}$

$10a^2 = 64b \quad \text{(Equation i)}$

The length of the major axis is $2b = 20$, so $b = 10$. (Equation ii)

Substitute Equation ii into Equation i:

$10a^2 = 64(10)$

$a^2 = 64$

$a = 8$

Substitute these values ($a^2 = 64$ and $b^2 = 100$) into the standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

$\frac{x^2}{64} + \frac{y^2}{100} = 1$

Example 2

Find the length of the latus rectum of an ellipse whose equation is $\frac{(x - 4)^2}{81} + \frac{(y - 3)^2}{16} = 1$.

Solution:

Given the equation $\frac{(x - 4)^2}{81} + \frac{(y - 3)^2}{16} = 1$.

Here, $a^2 = 81$ and $b^2 = 16$, so $a = 9$ and $b = 4$. Since $a>b$, the length of the latus rectum is $\frac{2b^2}{a}$:

$\frac{2(16)}{9} = \frac{32}{9} \text{ units.}$

The equation of an ellipse can also be written in parametric form. Consider the standard equation of the ellipse:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

This equation resembles the trigonometric identity $\cos^2\theta + \sin^2\theta = 1$.

From Figure 1.9, it can be observed that $x$ and $y$ are associated with cosine and sine, respectively.

To express the equation of an ellipse in parametric form, compare the terms of the two equations as follows:

$\frac{x^2}{a^2} = \cos^2\theta \quad \text{and} \quad \frac{y^2}{b^2} = \sin^2\theta.$

This implies that:

$x = a\cos\theta \quad \text{(Equation 1.13)}$

$y = b\sin\theta \quad \text{(Equation 1.14)}$

Thus, equations (1.13) and (1.14) are parametric equations of a standard ellipse with the center at the origin and foci on the x-axis. When the major axis is on the y-axis, or the center of the ellipse is at $(h,k)$, the parametric equations change slightly.

Example 1

Express the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ in parametric form.

Solution:

Given the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$. This implies that $a = 5$ and $b = 3$. The standard parametric equations of the ellipse are:

$x = a\cos\theta \quad \text{and} \quad y = b\sin\theta$

Substituting the values of $a$ and $b$ gives:

$x = 5\cos\theta \quad \text{and} \quad y = 3\sin\theta$

Example 2

The equation of the ellipse is given in parametric form as $x = -5 + 3\cos\theta$ and $y = 1 + 4\sin\theta$. Find its:

(a) Corresponding Cartesian equation.

(b) Length of the latus rectum.

Solution:

(a) Given $x = -5 + 3\cos\theta$ and $y = 1 + 4\sin\theta$.

Rearrange the equations:

$\cos\theta = \frac{x + 5}{3}$

$\sin\theta = \frac{y - 1}{4}$

Squaring and adding:

$\cos^2\theta + \sin^2\theta = \frac{(x + 5)^2}{9} + \frac{(y - 1)^2}{16}$

Since $\cos^2\theta + \sin^2\theta = 1$, the Cartesian equation is:

$\frac{(x + 5)^2}{9} + \frac{(y - 1)^2}{16} = 1$

(b) From the Cartesian equation, $a^2 = 9$ and $b^2 = 16$, so $a = 3$ and $b = 4$. Since b>a, the length of the latus rectum is $\frac{2a^2}{b} = \frac{2(9)}{4} = \frac{9}{2}$.

Example 3

The equation of the ellipse is given in Cartesian form as $9x^2 + 4y^2 + 18x - 24y + 9 = 0$. Find the corresponding parametric equations of the ellipse.

Solution:

Given $9x^2 + 4y^2 + 18x - 24y + 9 = 0$.

Rearranging and completing the square:

$9(x^2 + 2x) + 4(y^2 - 6y) = -9$

$9(x^2 + 2x + 1) + 4(y^2 - 6y + 9) = -9 + 9 + 36$

$9(x + 1)^2 + 4(y - 3)^2 = 36$

Divide by 36: $\frac{(x + 1)^2}{4} + \frac{(y - 3)^2}{9} = 1$

The center is $(-1, 3)$, $a^2 = 4$ so $a = 2$, and $b^2 = 9$ so $b = 3$.

The parametric equations of the translated ellipse are:

$x = h + a\cos\theta \quad \text{and} \quad y = k + b\sin\theta$

Substituting the values $h = -1$, $k = 3$, $a = 2$, and $b = 3$ gives:

$x = -1 + 2\cos\theta \quad \text{and} \quad y = 3 + 3\sin\theta$