# Solving Precal Problems: Ellipses & Hyperbolas

• vitaly
In summary, the conversation discusses two problems involving a hyperbola and an ellipse, where the goal is to write the equations of these shapes in standard form. The first problem involves finding the equation of a hyperbola, x^2 + 4xy + y^2 - 12 = 0, and the challenge is to get it into the form x^2/a^2 - y^2/b^2 = 1. The second problem involves finding the height of an arch at a specific point, and the solution involves using the standard equation for an ellipse, x^2/a^2 + y^2/b^2 = 1, and plugging in the given values. The conversation also mentions the possibility of
vitaly
Here are two problems that stumped our entire precal class. And we have a test soon, so I would like to be able to know how to work these type of problems.

1. Write the equation of the hyperbola, x^2 + 4xy + y^2 - 12 = 0, in standard form.
Okay, I know the formula needs to be x^2/a^2 - y^2/b^2 = 1, but I can't get it into that form... Is there a possibility that there is a typo. But, on the other hand, I checked to see if this is indeed a hyperbola, so I used the
B^2 - 4AC rule, and the result was greater than zero. That should mean the equation is a hyperbola or two intersecting lines. The problem is only getting it into standard form. Any ideas?

2. An arch in a cathedral has the shape of the top half an ellipse and is 40 feet wide and 12 feet high from the center from the floor. Find the height of the arch at 10 feet from the center?
I tried a lot of things, in desperation. Unfortunately, there are no examples in my book, and I couldn't find any online. I tried putting it into standard form of x^2/a^2 + y^2/b^2 = 1, but nothing worked. I can position the ellipse to be in the center, so a possible point would be (0,12), but I don't know where to go from there. All help is appreciated.

Have you tried completing the square for the first one?

If the equation contains both an x^2 and a y^2 i do not believe that would result in a parabola... if it is you've got me stumped.

Jameson said:
If the equation contains both an x^2 and a y^2 i do not believe that would result in a parabola... if it is you've got me stumped.

Where is a parabola mentioned in the question?

For question 2, you just need to pick values of a and b in the standard ellipse equation so that the ellipse has the correct shape (so that (0, 12) and (20, 0) are points on the ellipse). Then plug in the value of x that you are given and solve for the value of y.

Yes, I've tried completing the square. I got a strange answer, however. I'm just going to have to ask my teacher for help.

And thank you, master_coda, for help with number 2. The equation turned out to be 10^2/20^2 + y^2/12^2 = 1; where y = 10.4, which is the needed answer.

You can't get #1 into that form because it isn't possible. The equation does indeed define a hyperbola, but it is not aligned along the $x$ or $y$ axes, and thus it does not have the form you seek.

In fact, with a little knowledge of quadratic forms and some linear algebra, you can figure out that its semimajor axis is parallel to the vector $(1\ , \ 1)$ (and semiminor axis parallel to $(-1 \ , \ 1)$).

Using the transformation of coordinates $u = x+y$ and $v = x - y$ gives the equation

$$3u^2 - v^2 = 24 \Longleftrightarrow \frac{u^2}{8} - \frac{v^2}{24} = 1$$

for the hyperbola.

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## 1. What is the difference between an ellipse and a hyperbola?

An ellipse is a closed curve with two foci, while a hyperbola is an open curve with two branches that never intersect.

## 2. How do I find the center of an ellipse or hyperbola?

The center is located at the midpoint between the two foci for an ellipse, and at the intersection of the two asymptotes for a hyperbola.

## 3. What is the eccentricity of an ellipse or hyperbola?

The eccentricity is a measure of how "squished" or "stretched" the curve is. It is calculated by dividing the distance between the two foci by the length of the major axis for an ellipse, and by dividing the distance between the two vertices by the length of the transverse axis for a hyperbola.

## 4. How do I graph an ellipse or hyperbola?

To graph an ellipse or hyperbola, plot the center point and then use the length of the major and minor axes (for an ellipse) or the transverse and conjugate axes (for a hyperbola) to determine the vertices and foci. Then, draw the curve using the shape of the equation.

## 5. How do I use the foci and vertices to write the equation of an ellipse or hyperbola?

The general equation for an ellipse is (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k) is the center point and a and b are the lengths of the major and minor axes, respectively. The equation for a hyperbola is (x-h)^2/a^2 - (y-k)^2/b^2 = 1, where the transverse axis is the line connecting the two vertices and a and b are the distances from the center to the vertices and foci.

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