Plotting this point creates a triangle in quadrant I:

Using our knowledge of Special Right Triangles, we can conclude that the angle is and the radius/hypotenuse of this triangle is . Our polar coordinates are therefore , so we can eliminate that as a choice since we know it works.

Looking at the unit circle [or just the relevant parts] can give us a sense of what happens when the angles and/or the radii are negative:

Now we can easily see that the angle would correspond with our angle of , so  works.

We can see that if our radius is negative we'd want to start off at the angle , so the point  works.

As we can see from looking at this excerpt from the unit circle, another way of writing the angle would be to write , so the point  works.

The only one that does not work would be  because that would place us in quadrant II rather than I like we want.

The relation between polar coordinates and rectangular coordinates is given by  and .

You can plug in each of the choices for  and  and see which pair gives the rectangular coordinate .

The answer turns out to be .

Alternatively, you can find  by the equation 

, thus 

.

As for finding , you can use the equation

, and since  

.

Thus, the polar coordinate is . 

First, find r using pythagorean theorem,

Then we can find theta by doing the inverse tangent of y over x:

Since this point is in quadrant II, add 180 degrees to get

To convert from rectangular coordinates to polar coordinates :

Using the rectangular coordinates given by the question,

The polar coordinates are 

Start with the equation

It would be helpful if we "complete the square" for both the x and y polynomials.

So by adding 9 and 900 we transform the equation into

And then by dividing 900, we see that the outcome is:

The standard equation for an ellipse is

With a center at 

Therefore this is the equation of an ellipse, with center 

Remember that the equation for an ellipse in standard form looks like the following:

Where the point (h,k) gives the center of the ellipse, a is half the length of its axis in the x direction, and b is half the length of its axis in the y direction. We can see that this form has a 1 on the right side of the equation, so let's start by dividing both sides of our equation by 36 to get a 1 on the right side:

Now we can simplify the fractions on the right side of the equation, which gives us the equation for our ellipse in standard form:

This ellipse would have its center at (4,7), would be 6 units wide in the x direction, and 4 units wide in the y direction, because so , and so .

Remember that in order for the equation of an ellipse to be written in standard form, it must be written in one of the following two ways:

Where the point (h,k) gives the center of the ellipse, a is half the length of the axis for which it is the denominator, and b is half the length of the axis for which it is the denominator. Looking at our answer choices, we need one that involves addition, has different denominators under the x and y terms, and is equal to 1 on the right side of the equation. We can see that out of all the answer choices, the following is the only one that satisfies those requirements of standard form:

The equation of an ellipse in standard form is

,

where  is the center point,  is half the length of its axis in the  direction, and  is half the length of the axis in the  direction.

In this instance, the center point  , , and . Therefore:

Recall that the general equation of a circle is: 

. 

The key feature here being that both the  and  terms are squared.

Thus, we can narrow down the choices without this feature.

Next, consider the center. Since the center here is , we should not have any middle terms.

The only choice that fits that description is . 

The general equation for an ellipse is , although if we consider a to be half the length of the major axis, a and b might switch depending on if the longer major axis is horizontal or vertical. This general equation has as the center, a as the length of half the major axis, and b as the length of half the minor axis.

Because the center of this ellipse is at and the foci are at , we can see that the foci are away from the center, and they are on the horizontal axis. This means that the horizontal axis is the major axis, the one with length 14. Having a length of 14 means that half is 7, so . Since the foci are away from the center, we know that . We can solve for b using the equation :

that's really as far as we need to solve.

Putting all this information into the equation gives: