Ellipse Equation Exercises and Solutions

We want the sum of distances and to always equal , that is,

Therefore, we have:

If we isolate one radical, we obtain:

Then, squaring both sides, we have:

Notice that the term appears on both sides of the equation. Therefore, we can cancel it, so we have:

If we expand the two squared binomials, we will have:

Then, regrouping like terms—and dividing the equation by —we have:

We have eliminated one radical. To eliminate the other we repeat the procedure. We square the expression, expand the squared binomials, and regroup terms:

that is,

We know that the semi-major axis is the distance between the center and the vertex , that is,

Likewise, the semi-focal distance is the distance between the center and the focus of the ellipse—which is half the distance between the two foci—that is,

Finally, the semi-minor axis is calculated by:

Thus, the reduced equation of the ellipse is given by:

a)

We will describe the first part in detail. The others will be more summarized.

We know that the semi-major axis is the distance between the center and the vertex , that is,

Likewise, the semi-focal distance is the distance between the center and the focus of the ellipse—which is half the distance between the two foci—that is,

Finally, the semi-minor axis is calculated by:

Thus, the reduced equation of the ellipse is given by:

b)

We have:

Therefore, the semi-minor axis is given by:

Thus, the reduced equation of the ellipse is:

Notice that, in this case, we divide by instead of . This is because the major axis is vertical (note that , , and all have the same value in their coordinate).

c)

Notice that the coordinates of the three points are the same. Therefore, the major axis is vertical. Thus, we have:

Therefore, the semi-minor axis is given by:

Thus, the reduced equation of the ellipse is:

d)

Notice now that the coordinates are fixed at each point. Thus, the major axis of the ellipse will be horizontal. So, we have:

Furthermore,

Therefore, the reduced equation will be:

Observe the graph below:

We know that the focal distance must be . Thus, the semi-focal distance is:

which is the distance from the center to any focus. Thus, the distance between the center and the vertex is:

With this, we have:

Therefore, the equation of the ellipse is:

The equation of the ellipse must have the form:

because it has its center at the origin. Furthermore, we have that the ellipse passes through the point . Thus, it must satisfy:

If we solve for , we have . Then, since is greater than , it follows that:

Furthermore, in the eccentricity formula it must be satisfied that:

If we square the equation, it follows that:

We multiply the equation by , and then by to obtain:

By grouping like terms, we obtain:

That is, . Therefore, the equation of the ellipse is:

Since the ellipse has its center at the origin, then its equation must have the form:

Furthermore, since the minor axis measures , then the semi-minor axis is:

Then, since the ellipse passes through the point , it must satisfy the equation:

Solving for we have:

So that:

Thus, the equation of the ellipse is:

We have that the focal distance is . Therefore, the semi-focal distance is:

Likewise, the sum of the distances from any point to the foci is always constant. This distance coincides with the major axis, so:

Finally, the semi-minor axis measures:

Therefore, the equation of the ellipse is:

Since the ellipse passes through both points, it must satisfy the following system of equations:

This is a nonlinear system with two unknowns. In this case, we use a change of variable:

The solution to this system is:

To verify this, you can substitute the values of and into the original nonlinear system.

Therefore, the equation of the ellipse is:

The focal distance is . Therefore, we have:

Likewise, the semi-minor and semi-major axes satisfy—the relationship that , , and always fulfill:

On the other hand, we have a rectangle whose sides measure and . This rectangle has an area given by:

Therefore, we must solve the following nonlinear system of equations:

This nonlinear system can be solved by solving for from the second equation and substituting its value into the first equation. This yields a biquadratic equation. The solution to the system is given by:

Therefore, the equation of the ellipse is:

We know that the semi-minor axis is the distance between the center and the co-vertex , that is,

Likewise, the semi-focal distance is the distance between the center and the focus of the ellipse—which is half the distance between the two foci—that is,

Finally, the semi-major axis is calculated by:

Thus, the reduced equation of the ellipse is given by:

Semi-major axis:

We have , therefore, the semi-major axis is .
Semi-focal distance:
Here we have that the distance between the two foci is . Therefore, the semi-focal distance is .
Semi-minor axis:
We have where is the semi-minor axis. Thus,

So, the semi-minor axis measures .
Reduced equation:
Since we have the values of and , as well as the center—which is the midpoint of the foci, that is —then the reduced equation is given by:

Eccentricity:

Finally, the eccentricity of the ellipse is given by:

From the form of the equation, we can tell that the ellipse has its center at the origin. Furthermore, we have:

Thus the vertices have coordinates:

since the major axis lies on the -axis. The co-vertices are located at:

Likewise, we have that the semi-focal distance is:

Thus, the foci are located at:

Finally, the eccentricity is found by:

From the equation it immediately follows that the center is at . Likewise, the semi-minor and semi-major axes are:

Therefore,

Thus, the vertices are located at , that is:

Likewise, the foci are located at:

The co-vertices are located at the points:

a)

The center is at the origin. The semi-minor and semi-major axes are:

Thus, the vertices are located at:

The co-vertices are located at the points:

The semi-focal distance is:

So the foci are located at:

Finally, the eccentricity is:

b)

First we must write the equation in reduced form, so we divide by :

Then, from the equation it follows that the center is at the origin , and that:

So the vertices are located at:

and the co-vertices are located at:

Furthermore, the semi-focal distance is:

Thus, the foci are located at:

Finally, the eccentricity is given by:

c)

The equation is already in reduced form. From this equation we can see that the center is at . Furthermore, the semi-minor and semi-major axes are given by:

The semi-focal distance is given by:

Notice that the major axis lies on the -axis. Thus, the vertices are located at:

The co-vertices are located at:

And the foci are the points:

Finally, the eccentricity is given by:

d)

Finally, we have an equation that is not in reduced form. We first divide by to obtain:

From the equation we have:

Furthermore, notice that the major axis lies on the -axis. Thus, the vertices are located at:

The co-vertices are the points:

And the foci are located at:

Finally, the eccentricity is:

a)

To determine the important points of the ellipse, we must write its equation in reduced form. The way to do this is by completing the square:

Then, we divide by :

Thus, it is clear to see that the center is at . Furthermore, we can see that:

Likewise, the major axis is horizontal, so the vertices are located at:

The co-vertices are located at:

And the foci are located at:

b)

We complete the square again:

Then, we divide by :

Thus, it is clear to see that the center is at . Furthermore, we can see that:

Likewise, the major axis is vertical, so the vertices are located at:

The co-vertices are located at:

And the foci are located at:

c)

We complete the square again:

Then, we divide by :

Thus, it is clear to see that the center is at . Furthermore, we can see that:

Likewise, the major axis is horizontal, so the vertices are located at:

The co-vertices are located at:

And the foci are located at:

d)

We complete the square again:

Then, we divide by :

Thus, it is clear to see that the center is at . Furthermore, we can see that:

Likewise, the major axis is vertical, so the vertices are located at:

The co-vertices are located at:

And the foci are located at:

First observe the graph of the line and the ellipse:

From the figure we can deduce that we must find the coordinates of points and . Then, will be the midpoint of these two.

Finding the coordinates of and is equivalent to solving the nonlinear system of equations given by:

This system is also solved by substitution. The solutions are given by:

Therefore, the midpoint of the chord is given by:

Semi-minor axis:

We have , therefore, the semi-minor axis is .

Semi-focal distance:

Here we have that the distance between the two foci is . Therefore, the semi-focal distance is .

Semi-major axis:

We have where is the semi-major axis. Thus,

So, the semi-major axis measures .

Reduced equation:

Since we have the values of and , as well as the center—which is the midpoint of the foci, that is —then the reduced equation is given by:

Eccentricity:

Finally, the eccentricity of the ellipse is given by:

Semi-minor axis:

We have that the center is , the midpoint of the foci, therefore, the semi-minor axis is .

Semi-focal distance:

Here we have that the distance between the two foci is . Therefore, the semi-focal distance is .

Semi-major axis:

We have where is the semi-major axis. Thus,

So, the semi-major axis measures .

Reduced equation:

Since we have the values of and , as well as the center—which is the midpoint of the foci, that is —, then the reduced equation is given by:

Eccentricity:

Finally, the eccentricity of the ellipse is given by:

Semi-major axis:

We have that the center is , the midpoint of the foci, therefore, the semi-major axis is .

Semi-focal distance:

Here we have that the distance between the two foci is . Therefore, the semi-focal distance is .

Semi-minor axis:

We have where is the semi-minor axis. Thus,

So, the semi-minor axis measures .

Reduced equation:

Since we have the values of and , as well as the center—which is the midpoint of the foci, that is —, then the reduced equation is given by:

Eccentricity:

Finally, the eccentricity of the ellipse is given by:

Semi-major axis:

We have that the center is , therefore, the semi-major axis is .

Semi-focal distance:

Here we have that the distance from the center to the focus is .

Semi-minor axis:

We have where is the semi-minor axis. Thus,

So, the semi-minor axis measures .

Reduced equation:

Since we have the values of and , as well as the center, then the reduced equation is given by:

Eccentricity:

Finally, the eccentricity of the ellipse is given by: