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aksik [14]
4 years ago
8

Which is the equation of a hyperbola with directrices at y = ±2 and foci at (0, 4) and (0, −4)?

Mathematics
1 answer:
Advocard [28]4 years ago
4 0

Answer:

Step-by-step explanation:

First off, I'm assuming that when you said "directrices" you mean the oblique asymptotes, since hyperbolas do not have directrices they have oblique asymptotes.

If we plot the asymptotes and the foci, we see that where the asymptotes cross is at the origin. This means that the center of the hyperbola is (0, 0), which is important to know.

After we plot the foci, we see that they are one the y-axis, which is a vertical axis, which means that the hyperbola opens up and down instead of sideways. Knowing those 2 characteristics, we can determine that the equation we are trying to fill in has the standard form

\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1

We know h and k from the center, now we need to find a and b. Those values can be found from the asymptotes. The asymptotes have the standard form

y = ±\frac{a}{b}(x-h)+k

Filling in our asymptotes as they were given to us:

y = ±\frac{2}{1}(x-0)+0 where a is 2 and b is 1. Now we can write the formula for the hyperbola!:

\frac{(y-0)^2}{4}-\frac{(x-0)^2}{1}=1 which of course simplifies to

\frac{y^2}{4}-\frac{x^2}{1}=1

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Answer:

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I didn’t see any choices but an inverse function switches the x and y values.

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3 years ago
I need help please.....
maria [59]
Here, Function: h(t)= -16t² + 70t + 40

So, put the value of t, (time at which you want to calculate the height)

h(1) = -16(1)² + 70(1) + 40
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Now, h(2) = -16(2)² + 70(2) + 40
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In short, Your height depends on time, and at each time it would be different, can be expressed by the coordinates on a Graph: (1, 94) (2, 116) (3, 106)

Hope this helps!


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