Conic Sections - Pre-Calculus

The answer is because it is the only degree-2 polynomial with a negative leading coefficient.

Recall that the solution of a system of equations is given by the intersection points of the graphs. Thus, this question is really asking how many times a parabola and a line can intersect. Visualize a parabola and a line. For this purpose, let's say that the parabola is facing up. If a line is drawn horizontally under the vertex, then it would not intersect the parabola at all, so the system would have no solutions. If the line is tangent to, or just skims the edge, of the parabola then it would only intersect once and the system would have one solution. If the line goes straight through the parabola, then it would intersect twice. There is no other option for the orientation of the line and parabola. Thus, there cannot be three solutions.

The first step of the problem is to find the axis of symmetry using the following formula:

Where a and b are determined from the format for the equation of a parabola:

We can see from the equation given in the problem that a=1 and b=-3, so we can plug these values into the formula to find the axis of symmetry of our parabola:

Keep in mind that the vertex of the parabola lies directly on the axis of symmetry. That is, the x-coordinate of the axis of symmetry will be the same as that of the vertex of the parabola. Now that we know the vertex is at the same x-coordinate as the axis of symmetry, we can simply plug this value into our function to find the y-coordinate of the vertex:

So the vertex occurs at the point:

Find the axis of symmetry and the vertex of the parabola given by the following equation:

To find the axis of symmetry of a parabola in standard form, , use the following equation:

So...

This means that we have an axis of symmetry at . Or, to put it more plainly, at we could draw a vertical line which would perfectly cut our parabola in half!

So, we are halfway there, now we need the coordinates of our vertex. We already know the x-coordinate, which is 7. To find the y-coordinate, simply plug 7 into the parabola's formula and solve!

This makes our vertex the point

Group the x and y terms on one side of the equation and the constant on the other.

Complete the square by taking half of the middle number for each variable and squaring it. Add the number to the other side of the equation.

Recall the standard form of the equation of a vertical parabola:

, where is the vertex of the parabola and gives the focal length.

When , the parabola will open up.

When , the parabola will open down.

For the parabola in question, the vertex is and . This parabola will open up. Because the parabola will open up, the directerix will be located unit down from the vertex. The equation for the directerix is then .

To solve for x and y, set both equations equal to each other and solve for x.

Substitute into either parabola.

The coordinate of intersection is .

Set both parabolas equal to each other and solve for x.

Substitute both values of into either parabola and determine .

The coordinates of intersection are:

and

To solve, set both equations equal to each other:

To solve as a quadratic, combine like terms by adding/subtracting all three terms from the right side to the left side:

This simplifies to

Solving by factoring or the quadratic formula gives the solutions and .

Plugging each into either original equation gives us:

Our coordinate pairs are and .

To solve, set the two quadratics equal to each other and then combine like terms:

subtract everything on the right from both sides to combine like terms.

Solving by factoring or using the quadratic formula gives us the solutions and .

To find the y-coordinates, plug these into either equation:

To solve, first re-write the second one so that y is isolated on the left side:

Now set the two quadratics equal to each other:

add/subtract all of the terms from the right side so that this is a quadratic equal to zero.

combine like terms.

Using the quadratic formula or by factoring, we get the two solutions and .

To get the y-coordinates, plug these numbers into either function:

To solve, set the two polynomials equal to each other:

add/subtract all of the terms from the right side from both sides.

combine like terms.

Solving with the quadratic formula or by factoring gives us the solutions 5 and -3.

To get the y-coordinates, plug these numbers into either of the original equations:

To determine the polar equation, first we need to interpret the original cartesian graph. This is an ellipse with a vertical major axis with half its length . The minor axis has half its length . To find the foci, use the relationship

so

Since the center is , this means that the foci are at

Having a focus at the origin means we can use the formula \[or sine\] where e is the eccentricity, and for an ellipse .

In this case, the focus at the origin is above the directrix, so we be subtracting. The major axis is vertical, so we are using sine.

Solving for p gives us:

The eccentricity is , in this case

This gives us an equation of:

We can simplify by multiplying top and bottom by 2:

To convert this cartesian equation to polar form, we will use the substitutions and .

First, we should expand the expression:

square x - 3

subtract 9 from both sides

group the squared variables next to each other - this will help see how to re-write this in polar form:

now make the substitutions:

This is a quadratic in r with

Solve using the quadratic formula:

Subtracting which would give us a circle with no radius, but adding so our answer is:

The directrix is to the left of the focus at the origin, and the major axis is horizontal, so our equation is going to take the form

where e is the eccentricity. For a hyperbola specifically,

First we need to solve for c so we can find the eccentricity. For a hyperbola, we use the relationship where is half the length of the minor axis and is half the length of the major axis, in this case the horizontal one.

In this case, and

add 9 to both sides

take the square root

To find the eccentricity: so in this case

gives us

Plugging in e and p:

To simplify we can multiply top and bottom by 3:

To put this in polar form, we need to understand its structure. We can find the foci by using the relationship where a is half the length of the major axis, and b is half the length of the minor axis.

In this problem,

.

Plugging these values into the above realationship we can solve for .

divide by -1

Since the center is , the foci are at and .

The major axis is vertical, and the lower focus is at the origin, so the polar equation will be in the form where and for an ellipse, .

Here, and

Simplify the top by cancelling out

Simplify by multiplying top and bottom by

Before writing this in polar form, we need to know the structure of this hyperbola. We can determine the foci by using the relationship

In this case, subtract 27

The center is at and the major axis is horizontal, so the foci are and , adding or subtracting 6 from the x-coordinate. Because the left focus is at the origin and the major axis is horizontal, our polar equation will be in the form where and for a hyperbola

Here,

Plugging this in gives:

simplify the numerator by cancelling the 2's

simplify once more by multiplying top and bottom by

This is the equation for a parabola, so the eccentricity is 1. It opens up, so the focus is above the directrix.

This means that our equation will be in the form

where a is the distance from the focus to the vertex.

In this case, we have , so .

Because the vertex is , the focus is so we can use the formula without adjusting anything.

The equation is

.

This is the equation for a parabola with a vertex at . This parabola opens left because it is negative. The distance from the focus to the vertex can be found by solving , so . This places the focus at the origin.

Because of this and the fact that the focus is to the left of the directrix, we know that the polar equation is in the form

.

That means that this equation is

.

Recall the standard form of the equation of a horizontal parabola:

, where is the vertex of the parabola and is the focal length.

When , the parabola opens to the right.

When , the parabola opens to the left.

For the given parabola, the vertex is and . This means the parabola is opening to the left and that the directerix will be located units to the right of the vertex. The directerix is then .