Exercises and Problems on the Parabola Equation

We will proceed by determining, in reduced form, the equations of the parabolas, indicating the value of the parameter , and with this the coordinates of the focus and the equation of the directrix.

1

We solve for the quadratic term:

We identify the value of p:

We locate the focus and find the equation of the directrix:

Finally, we graph using the data obtained.

2

We solve for the quadratic term:

We identify the value of p:

We locate the focus and find the equation of the directrix:

Finally, we graph using the data obtained.

3

We solve for the quadratic term:

We identify the value of p:

We locate the focus and find the equation of the directrix:

Finally, we graph using the data obtained.

We will again proceed by determining, in reduced form, the equations of the parabolas, indicating the value of the parameter , and with this the coordinates of the focus and vertex.

1

We complete the perfect square trinomial and solve for it:

We factor:

With the equation we identify its elements:

With the vertex and the value of the parameter , we locate the focus and directrix:

Finally, we locate on the graph.

2

We complete the perfect square trinomial and solve for it:

We factor:

With the equation we identify its elements:

With the vertex and the value of the parameter , we locate the focus and directrix:

Finally, we locate on the graph.

3

We complete the perfect square trinomial and solve for it:

We factor

With the equation we identify its elements:

With the vertex and the value of the parameter , we locate the focus and directrix:

Finally, we locate on the graph.

We can express the equation of the parabola as follows:

From this we deduce that:

Therefore, the focus has coordinates and the equation of the directrix is .

The latus rectum of a parabola is the chord drawn through the focus that is parallel to its directrix. To calculate the length of the latus rectum, we calculate the value of "" for . Thus, if , then:

Here, we have taken the positive value since we are talking about distance.

Thus, the length of the latus rectum is twice this distance, that is:

a) By locating the directrix and focus, it is easy to deduce that the parabola opens to the right and its vertex is at the origin.

Knowing that the focus for these parabolas has coordinates latex[/latex], we conclude that:

The equation is:

b) By locating the directrix and focus, it is easy to deduce that the parabola opens upward and its vertex is at the origin.

Knowing that the focus for these parabolas has coordinates latex[/latex], we conclude that:

We substitute in the equation:

c) By locating the directrix and focus, it is easy to deduce that the parabola opens to the left and its vertex is at the origin.

The focus for parabolas that open to the left has coordinates , which means that:

We substitute in the equation:

d) By locating the directrix and vertex, it is easy to deduce that the parabola opens downward and its vertex is at the origin.

For these parabolas the equation of the directrix is . Therefore:

The equation is of the form:

a) By locating the focus and vertex, it is easy to deduce that the parabola opens to the right and its vertex is at the origin. Therefore, its equation is of the form:

Recall that for these parabolas, the focus is located at , therefore:

Finally, the parabola has an equation of the form:

b) By locating the vertex and focus, we can note that the focus is to the left of the vertex, which indicates that the parabola opens to the left and its equation is of the form:

We calculate the distance from the vertex to the focus and obtain:

We substitute in the equation:

c) By locating the vertex and focus, we can note that the focus is above the vertex, which indicates that the parabola opens upward and its equation is of the form:

We calculate the distance from the vertex to the focus and obtain:

We substitute in the equation:

d) By locating the vertex and focus, we can note that the focus is to the right of the vertex, which indicates that the parabola opens to the right and its equation is of the form:

We calculate the distance from the vertex to the focus and obtain:

We substitute in the equation:

We know that the distance from a point to the focus equals the distance from that point to the directrix.

The distance from a line to a point is given by:

Thus, if we consider the vertex (which we do not know) as the point , the first equation is equivalent to:

We square both sides to eliminate the square root on the left and develop:

We simplify, leaving variables on one side, and on the other:

We factor:

Given that the axis is horizontal and its vertex is at the point , we use the equation:

Substituting the vertex coordinates:

If the parabola passes through the point , then:

Thus, we have the equation:

Simplifying the equation above we have that the equation of the parabola we're looking for is:

The equation of a vertical parabola is of the form:

Since points A, B, and C lie on the parabola, their coordinates satisfy its equation:

We solve the system of equations and obtain:

Thus, the equation of the parabola is:

To calculate the relative position between both objects, we need to see if intersection points exist. The coordinates of such points should satisfy both equations.

To solve the system, we substitute the second equation into the first:

We expand:

We solve the quadratic using the quadratic formula:

We have the coordinates; to obtain the coordinates, we substitute in one of the equations. In this case, the simplest is:

Therefore:

Thus, the intersection points are:

Therefore, the line is secant to the parabola, as shown in the following graph.

To calculate the relative position between both objects, we need to see if intersection points exist. The coordinates of such points should satisfy both equations.

To solve the system, we equate both equations and solve for :

Using the quadratic formula to solve the equation above, we obtain:

We have the coordinates. To obtain the coordinates, we substitute in one of the original equations. We substitute in the second equation:

Thus, the intersection points are:

Therefore, the line is secant to the parabola, as shown in the following graph.