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Which of the following is a foci of the conic section with the equation x^2/16-y^2/4=1

User Roy Miloh
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2 Answers

4 votes

Answer: See below

Step-by-step explanation:The foci of a conic section can be found using the formula c = √(a^2 + b^2), where a and b are the coefficients of x^2 and y^2, respectively.

In the equation x^2/16 - y^2/4 = 1, the values of a and b are 4 and -2, respectively.

Using the formula, we can calculate c as follows:

c = √(4^2 + (-2)^2)

= √(16 + 4)

= √20

= 2√5

So, the value of c is 2√5.

The foci of the conic section are located on the transverse axis, which is the x-axis in this case. The distance from the center to each focus is given by the value of c.

Since the center of the conic section is (0, 0), the foci can be found by adding or subtracting the value of c to the x-coordinate of the center.

Comparing the given options, we can see that (0, 2√5) matches the coordinates of one of the foci.

Therefore, the correct answer is (0, 2√5), which is one of the foci of the conic section with the equation x^2/16 - y^2/4 = 1.

User Logical Chimp
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5.4k points
6 votes

Answer:

I think the answer is A. (2√5, 0)

Explanation:

Sorry if i am wrong :(

User Alex Guerra
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5.6k points