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Which of the following is a foci of x2/16-y2/4 ? (2√5, 0) (0, 2√5) (2√3, 0)

User Pills
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2 Answers

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Answer: (2√5, 0)

Step-by-step explanation: got it right

User Helgatheviking
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3 votes

foci of
(x^(2))/(16) -(y^(2))/(4) =1 is
(2√(5), 0). correct option a.

Explanation:

Complete equation of Hyperbola for the above question is x2/16-y2/4 = 1 or
(x^(2))/(16) -(y^(2))/(4) =1 , Simplify each term in the equation in order to set the right side equal to 1 . The standard form of an ellipse or hyperbola requires the right side of the equation be 1 . This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola:


((x-a)^(2))/(a^(2)) - ((y-k)^(2))/(b^(2)) = 1

a=4 , b = 2, h = 0 , k= 0

Find the distance from the center to a focus of the hyperbola by using the following formula:


\sqrt{a^(2)+b^(2)}

Substitute the value of a and b in the formula.


\sqrt{4^(2)+2^(2)} =
2√(5)

The first focus of a hyperbola can be found by adding c to h

(h+c,k)

Substitute the known values of h , c , and k into the formula and simplify:


(2√(5), 0) . ∴ foci of
(x^(2))/(16) -(y^(2))/(4) =1 is
(2√(5), 0). correct option a.

User Drew Turner
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8.2k points

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