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anonymous
 5 years ago
(x/2) squared  (y/2) squared=1
find the asymptotes
anonymous
 5 years ago
(x/2) squared  (y/2) squared=1 find the asymptotes

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let’s first consider something easier. x^2 – y^2 = 1. This will trace out a hyperbola. The asymptotes are lines that the hyperbola will approach as x goes to infinity. Asymptotes are lines. So we want equations of lines. This suggests we solve for y in terms of x in the hyperbola, and then see what happens to y as x goes to infinity. Our hyperbola is x^2 – y^2 = 1. So y^2 = x^2 – 1. So y = sqrt(x^2 – 1) or y =  sqrt(x^2 – 1). Now, we see that when x goes to infinity, the expression under the square root goes to infinity. So y = sqrt(x^2 – 1) goes off to +infinity and y =  sqrt(x^2 – 1) goes off to – infinity. But what line does it approach? As x becomes large, x^2 – 1 is approximately just x^2. So then y is approximately sqrt(x^2) = x. Therefore, y = x is an asymptote. Similarly, for negative values of y, y =  sqrt(x^2 – 1) means that y is approximately –sqrt(x^2) = x. So y = x and y = x are the asymptotes. In general, if your hyperbola is (x/a)^2 – (y/b)^2 = 1, then (y/b)^2 = (x/a)^2 – 1. So y/b = sqrt( (x/a)^2 – 1) or y/b =  sqrt((x/a)^2 – 1). So as x goes to infinity, (x/a)^2 – 1 is approximately just (x/a)^2 since subtracting 1 becomes insignificant. So then y/b is approximately sqrt( (x/a)^2) = x/a. Therefore, y ~ (b/a) * x. Similarly for the negative values of y, y ~ (b/a) * x. These are your asymptotes for the hyperbola (x/a)^2 – (y/b)^2 = 1. You can also use the same chain of reasoning to find the asymptotes for the hyperbola (y/b)^2 – (x/a)^2 = 1 if you needed to.
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