Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conic Sections

  1. Apr 6, 2005 #1
    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 possibilty 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.
  2. jcsd
  3. Apr 6, 2005 #2
    Have you tried completing the square for the first one?
  4. Apr 6, 2005 #3
    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.
  5. Apr 6, 2005 #4
    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.
  6. Apr 6, 2005 #5
    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.
  7. Apr 6, 2005 #6
    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 [itex]x[/itex] or [itex]y[/itex] 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 [itex](1\ , \ 1)[/itex] (and semiminor axis parallel to [itex](-1 \ , \ 1)[/itex]).

    Using the transformation of coordinates [itex]u = x+y[/itex] and [itex]v = x - y[/itex] gives the equation

    [tex]3u^2 - v^2 = 24 \Longleftrightarrow \frac{u^2}{8} - \frac{v^2}{24} = 1[/tex]

    for the hyperbola.
    Last edited: Apr 6, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook