Select the correct answer. which hyperbola has both foci lying in the same quadrant? a. b. c. d.
1 answer:
The hyperbola having both foci lying in the same quadrant is <u>(y - 16)²/9² - (x + 1)²/12² = 1</u>, making the <u>4th option</u> a right choice.
For the hyperbola with an equation of the form (x - h)²/a² - (y - k)²/b² = 1, the foci in on the points (h + c, k) and (h - c, k), where c = √(a² + b²).
For the hyperbola with an equation of the form (y - k)²/a² - (x - h)²/b² = 1, the foci in on the points (h, k + c) and (h, k - c), where c = √(a² + b²).
<u>Among the options</u>:
(a) (x - 24)²/24² - (y -1)²/7² = 1.
The equation is of the form (x - h)²/a² - (y - k)²/b².
h = 24, k = 1, a = 24, b = 7.
c = √(a² + b²) = √(24² + 7²) = √(576 + 49) = √625 = 25.
Thus, foci are at the points (h + c, k), (h - c, k) = (24 + 25, 1), (24 - 25, 1) = (49, 1), and (-1, 1), which are in different quadrants.
(b) (y - 12)²/5² - (x - 6)²/12² = 1.
The equation is of the form (y - k)²/a² - (x - h)²/b².
h = 6, k = 12, a = 5, b = 12.
c = √(a² + b²) = √(5² + 12²) = √(25 + 144) = √169 = 13.
Thus, foci are at the points (h, k + c), (h, k - c) = (6, 12 + 13), (6, 12 - 13) = (6, 25), and (6, -1), which are in different quadrants.
(c) (y - 16)²/15² - (x - 2)²/8² = 1.
The equation is of the form (y - k)²/a² - (x - h)²/b².
h = 2, k = 16, a = 15, b = 8.
c = √(a² + b²) = √(15² + 8²) = √(225 + 64) = √289 = 17.
Thus, foci are at the points (h, k + c), (h, k - c) = (2, 16 + 17), (2, 16 - 17) = (2, 33), and (2, -1), which are in different quadrants.
(d) (y - 16)²/9² - (x + 1)²/12² = 1.
The equation is of the form (y - k)²/a² - (x - h)²/b².
h = -1, k = 16, a = 9, b = 12.
c = √(a² + b²) = √(9² + 12²) = √(81 + 144) = √225 = 15.
Thus, foci are at the points (h, k + c), (h, k - c) = (-1, 16 + 15), (-1, 16 - 15) = (-1, 31), and (-1, 1), which are in the same quadrants (QUADRANT II).
Thus, the hyperbola having both foci lying in the same quadrant is <u>(y - 16)²/9² - (x + 1)²/12² = 1</u>, making the <u>4th option</u> a right choice.
Learn more about the foci of a hyperbola at
#SPJ4
For the options, refer the image.
You might be interested in
It might help to draw out a probability tree. Check out the diagram below. The first two branches at the top represent bags A and B. The probability of picking either bag is 1/2 assuming each bag is equally likely to be picked.
If we go with bag A, then the probability of getting blue is 3/11 since there are 3 blue out of 8+3 = 11 total. The probability of getting bag A and a blue marble is (1/2)*(3/11) = 3/22 which I've marked in the diagram as well.
If we go with bag B, then the probability of getting blue is 7/12 since there are 7 blue out of 5+7 = 12 total. The probability of getting bag B and blue is (1/2)*(7/12) = 7/24
Add up the results found getting
P(blue) = P(blue & bag A) + P(blue & bag B)
P(blue) = 3/22 + 7/24
P(blue) = 36/264 + 77/264
P(blue) = 113/264
This is shown in the diagram as well.
--------------------------------------------------
Recall that conditional probability in general is defined as follows
P(A given B) = P(A and B)/P(B)
In this problem, we'll have
P(bag B given blue) = P(bag B and blue)/P(blue)
Therefore, we can say...
P(bag B given blue) = P(bag B and blue)/P(blue)
P(bag B given blue) = (7/24)/(113/264)
P(bag B given blue) = (7/24)*(264/113)
P(bag B given blue) = (7*264)/(24*113)
P(bag B given blue) = (7*24*11)/(24*113)
P(bag B given blue) = (7*11)/(113)
P(bag B given blue) = 77/113
In short, the probability of getting bag B, given the marble is blue, is 77/113.
