Question

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

Answer 1

Answer: (2√5, 0)

Step-by-step explanation: got it right

Answer 2

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.