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Poznerrr
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Member warned to not delete the homework template in future posts
I have tried like 5 times to do this problem ans still don't get the answer I'm supposed to (A). Anybody finds mistake in my work below?
Yeah, I thought about it, and it works out with a/b being the slope of the asymptote. But different sources stick to either a/b and b/a, so it's really confusing which slope is actually correct.Fightfish said:If I'm not mistaken, it should be ##a/b## and not ##b/a##. so the factor appearing will be ##2/3## and not ##3/2##.
The way to remember it is well, to derive it yourself, which isn't very hard. The general equation for a hyperbola is in the formPoznerrr said:Yeah, I thought about it, and it works out with a/b being the slope of the asymptote. But different sources stick to either a/b and b/a, so it's really confusing which slope is actually correct.
micromass said:Given a conic section
[tex]Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0[/tex]
the asymptotes can be calculated by a little trick. Basically, the slope of the asymptote is ##s##, where ##s## is the solution to ##A + Bs + Cs^2=0##.
In your case, this corresponds to the solution of ##4-9s^2=0##. So we get ##s = 2/3## or ##s= -2/3##. The slope of (A) is ##-2/3##.
Thanks! I'll use this trick in future.micromass said:Given a conic section
[tex]Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0[/tex]
the asymptotes can be calculated by a little trick. Basically, the slope of the asymptote is ##s##, where ##s## is the solution to ##A + Bs + Cs^2=0##.
In your case, this corresponds to the solution of ##4-9s^2=0##. So we get ##s = 2/3## or ##s= -2/3##. The slope of (A) is ##-2/3##.
Poznerrr said:Thanks! I'll use this trick in future.
Poznerrr said:Thanks! I'll use this trick in future.
A hyperbola's asymptote is a straight line that the hyperbola approaches but never touches. It is used to describe the shape of the hyperbola and its behavior as its distance from the center increases.
To find the asymptotes of a hyperbola, you can use the equation y = mx + b, where m is the slope of the asymptote and b is the y-intercept. The slope of the asymptotes can be found by dividing the coefficient of x by the coefficient of y in the hyperbola's equation.
A vertical asymptote is a line that is parallel to the y-axis and approaches the hyperbola as x approaches a certain value. A horizontal asymptote is a line that is parallel to the x-axis and approaches the hyperbola as y approaches a certain value. The difference lies in the direction in which the asymptote approaches the hyperbola.
No, a hyperbola can only have two asymptotes. This is because a hyperbola is defined as a curve with two distinct branches, and each branch can have at most one vertical and one horizontal asymptote.
People commonly make mistakes when finding the slope of the asymptotes or determining the direction of the asymptotes. It is important to remember that the slope of a vertical asymptote is undefined, and the direction of the asymptotes can be determined by looking at the signs of the coefficients in the hyperbola's equation.