Reminder:

Ellipses have two perpendicular axes. Ellipses is symmetric on each of that axes. Larger of axes, which corresponds to distance beetwen line segment from $ -a$ to $a$ we called major axis. Smaller  axis, which corresponds to distance beetwen line segment from $ -b$ to $b$ we called minor axis.

The semi-major axis (labeled by letter $a$) and the semi-minor axis (labeled by $b$ in the figure) has half of the value of the major and minor axes. Whe ussually called that axes major and minor radius.

Two points $F_1$ and $F_2$ we usually called focal points. The sum of of distances prom any point $P$ and points $F_1$ and $F_2$ is constant and its equal to double value of major axis $a$.

$|PF_1|+|PF_2|=2a$

The distance from focal point to the cenete or the ellipse we called linear eccentricity and we labeled with letter $f$.

The eccentricity of an ellipse we usually labeled by letter $e$. Value of $e$ is the ratio of distances betwen to focal points and the length of the major axis.

$e=\frac{f}{a}$

The value of the eccentricity for an ellepse is between zero and one.

$0<e<1$

If the value of eccentricity  $e$ is one, that ellipse is circle. So, the circle is special case of ellipse for:

$e=1$.

Line and ellipse

Line and ellipse can be found in three different relation:

• they can intersect in two points,
• they can have one point in common,
• or their intersection can be an empty set.

Now that we know about their relations graphically, it would be good to know how to find them out using calculations.

We can determine the intersection of an ellipse and a line by solving the system of two following equations.

$ y = kx + l$

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

By inserting the equation of the line into an equation of the ellipse we get a quadratic equation. Using its discriminant we can find out in which relation are the ellipse and the line.

If the discriminant of that quadratic equation is equal to zero, the line is the tangent of an ellipse, if the discriminant is a negative number, line and ellipse have no points in common, and if the discriminant is greater than zero, this means that we can get two real solutions and that we have two intersections.

Example 3. Find the intersections of an ellipse $ x^2 + 9y^2 = 255$ and a line $ y = \frac{1}{3}x – 1$.

First thing you notice is that ellipse is looking a bit strange. There are no fractions and the right side is not equal to $1$. There is an easy way to fix this look in case you need it. Simply divide by “anything” that is on the left side. In this example you may also have few worries with that 9 that multiplies $ y^2$. Here you’d simply take that number $9$ and put it below “255” like denominator. Since we know how complex fractions are solved, this is no problem.

But, notice that you don’t have to do this. Since you are looking for the intersection, you have to solve the system. You can do this, even more easily, without transforming ellipse into its general form.

$ x^2 + 9y^2 = 255$

$ x^2 + 9(\frac{1}{3}x – 1)^2 = 255$

$ x^2 + 9(\frac{1}{9}x^2 – \frac{2}{3} + 1) = 255$

$ x^2 + x^2 – 6x + 9 = 255$

$ 2x^2 + 3 – 216 = 0$

The solutions of this quadratic equation are $ x_1 = 12$, and $ x_2 = – 9$ which means that $ y_1 = 3$, $ y_2 = – 4$.

We can check our solution using a sketch.

From the quadratic equation we can also derive the condition of a line to touch an ellipse.

A line touches an ellipse if and only iff the discriminant is equal to $0$.

Using a general equations of an ellipse and a line we get the following system:

$ y = kx + l$

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

$\frac{x^2}{a^2} + \frac{(kx + l)^2}{b^2} = 1$

$ (a^2k^2 + b^2)x^2 + 2kla^2x + a^2(l^2 – b^2) = 0$

$ D = 0 \rightarrow (2kla^2)^2 – 4(a^2k^2 + b^2)a^2(l^2 – b^2) = 0$

After the multiplication and division with $ 4a^2b^2$ we get to a condition $ a^2k^2 + b^2 = l^2$.

The condition of a line to touch an ellipse:

$ y = kx + l$ 

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

$ a^2k^2 + b^2 = l^2$

Example 4. From the point T(10, 5) draw tangents on an ellipse $ x^2 + 4y^2 = 40$

In these kinds of tasks it’s always smart to transform an equation into the general one, because from it you can clearly see what are a and b.

$\frac{x^2}{40} + \frac{y^2}{10} = 1$

$ a^2 = 40, b^2 = 10$

Which means that from $ a^2k^2 + b^2 = l^2$ we know that $ 40k^2 + 10 = l^2$. Here we have an equation with two unknowns. We can’t solve it without any more information. We can get all the information we need from the fact that those tangents go through the point $T(10, 5)$.

$ y = kx + l \rightarrow 5 = 10k + l$

Now, we have two equations with two unknowns.

From the second equation we can see that $ l = 5 – 10k$. By inserting this equation into the first one we get:

$ 40k^2 + 10 = 25 – 100k + 100k^2$

$ 12k^2 – 20k + 3 = 0$

$ k_1 = \frac{3}{2}, k_2 = \frac{1}{6}$. $ l_1 = – 10, l_2 = \frac{10}{3}$

From here we can write the equations of the tangents:

$ y = \frac{3}{2}x – 10$, $ y = \frac{1}{6}x + \frac{10}{3}$