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Murljashka [212]
2 years ago
13

Which is the equation of a hyperbola with directrices at x = ±3 and foci at (4, 0) and (−4, 0)?​

Mathematics
1 answer:
Alex Ar [27]2 years ago
5 0

Answer:

\frac{x^2}{12}-\frac{y^2}{4}=1

Step-by-step explanation:

Since our foci are located on the x-axis, then our major axis is going to be the horizontal transverse axis of the hyperbola:

<u>Formula for hyperbola with horizontal transverse axis centered at origin</u>

  • <u />\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
  • Directrices -> x=\pm\frac{a^2}{c}
  • Foci -> (\pm c,0) where a^2+b^2=c^2
  • a>b

Since we are given our directrices of x=\pm3 and foci of (\pm4,0), then we can set up the directrices equation to solve for a^2:

x=\pm\frac{a^2}{c}\\ \\\pm3=\pm\frac{a^2}{4}\\ \\12=a^2

Now we can determine b^2 to complete our equation for the hyperbola:

a^2+b^2=c^2\\\\12+b^2=4^2\\\\12+b^2=16\\\\b^2=4

Therefore, our equation for our hyperbola is \frac{x^2}{12}-\frac{y^2}{4}=1

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The question is in the screenshot attached. Thank you!
Rzqust [24]

Answer:

Domain: -4 < x < 4

Zeros: (-2, 0), (0, 0) and (2, 0)

The function is positive if: 0 < x < 2

The function is negative if: -4 < x < 0 and 2 < x < 4

Step-by-step explanation:

Domain of the function are those x values where the function is defined, For this case, -4 < x < 4

Zeros of a function are those x values where y = 0, that is, the graph intersect x-axis. For this case, the points are: (-2, 0), (0, 0) and (2, 0)

The function is positive if the graph o the function is above x-axis. For this case, the function is positive at the interval (0, 2)

The function is negative if the graph o the function is below x-axis. For this case, the function is negative at the intervals (-4, 0) and (2, 4)

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<u>Step-by-step explanation:</u>

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