Math II: Asymptote question!

<p>Hello,</p>

<p>I came across this question in one of the model tests in the Barron's book:</p>

<p>Q: Find the horizontal asymptotes of 4x^2 - 9y^2 = 36.</p>

<p>The solution in the book involves simplifying of the equation to that of a hyperbola and then using the formula for asymptotes of a hyperbola.</p>

<p>However, initially I did not recognise that I could reduce it to the equation of a hyperbola. I was trying to solve the question using the knowledge I have of horizontal asymptotes, which is:
1. If n>m there is no asymptote.
2. If n = m+1, there is a slant asymptote, in which case we use divide num by denom and the quotient is the asymptote.
3. IF n<m, it is y = 0
4. IF n = m, asymptote is ratio of leading coefficients.</p>

<p>(where n and m are degrees of num and denom respectively)</p>

<p>Is it possible to solve the given question using these facts? If yes, how?</p>

<p>Another question I had a problem with was:</p>

<p>Q. The intersection of a plane and a cylinder can be which of the following?</p>

<p>I. Circle
II. Parallel lines
III. Intersecting lines</p>

<p>The answer is I and II. However, how can the intersection represent a pair of lines? It will always be a planar figure, won't it?</p>

<p>You’re probably thinking of how to find horizontal asymptotes of a rational function in the form y = p(x)/q(x), where p(x) and q(x) are polynomials of degree n, m, respectively. You can’t use that here for two reasons, 1) we’re not dealing with a rational function in terms of one variable and 2) the asymptotes have slope not equal to 0.</p>

<p>Q1: The hyperbola can be written as (x^2)/9 - (y^2)/4 = 1. Clearly, the center of the hyperbola is (0,0). In general, the slope of the asymptote of a hyperbola in the form (x^2)/(a^2) - (y^2)/(b^2) = 1 is +/- b/a. In this case, the slope of the asymptote is +/- 2/3. Both asymptotes go through the point (0,0), so the equations of the asymptote are y = +/- 2x/3.</p>

<p>Q2: Not entirely positive on this, but I’m fairly certain a cylinder doesn’t have to have finite dimensions. For example, the cylindrical-coordinate graph of r = 1 in 3-space will be a cylinder. In this case, we can intersect a plane (e.g. x = 0) and the result will be two parallel lines.</p>

<p>But then why am I unable to use rule #2, which is the method to find slant asymptotes?</p>

<p>Also, is there some standard method for finding asymptotes of equations of this kind, or should I be able to identify that it is a hyperbola and then use the formula for its asymptotes?</p>

<p>For Q2, even if the cylinder does not have finite dimensions, the intersection you described will give us a pair of parallel planes and not lines, right?</p>

<p>Still, that’s only for explicitly defined rational functions, such as y = (5x^2 + 1)/(3x - 5). The hyperbola equation here is an implicit function. I would just use the +/- b/a above. The standard method to find the slope of the asymptote is to find the derivative of the given function, then take the limit as x approaches infinity (or negative infinity).</p>

<p>For Q2, it will be a pair of lines. Think of a cylinder of radius 1, infinitely long (e.g. r = 1 in cylindrical). If you pass a plane (x = 0) through the cylinder, the intersection of that plane with the surface will be two parallel lines.</p>

<p>Just in case you’re interested:</p>

<p>y = (5x^2 + 1)/(3x - 5) (rational function)
dy/dx = ((10x)(3x - 5) - (5x^2 + 1)(3))/((3x - 5)^2) = (15x^2 - 50x - 3)/((3x - 5)^2)</p>

<p>The limit as x approaches infinity of dy/dx is 15/9 or 5/3, so the slope of the asymptote is 5/3. //</p>

<p>Now consider the hyperbola (x^2)/9 - (y^2)/4 = 1. Taking dy/dx of both sides,
2x/9 - (y/2)(dy/dx) = 0 → dy/dx = 4x/9y</p>

<p>For very large (x,y), (x^2)/9 ≈ (y^2)/4, so x/y ≈ 3/2. Therefore, for large (x,y), dy/dx approaches 4(3)/9(2) = 12/18 = 2/3, which is equal to b/a in the above formula. Of course if we do this same logic on the other side we get -2/3.</p>