Question

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

Answer 1

foci of $\frac{x^{2}}{16} -\frac{y^{2}}{4} =1$ is $(2\sqrt{5}, 0)$. correct option a.

Step-by-step explanation:

Complete equation of Hyperbola for the above question is x2/16-y2/4  = 1 or $\frac{x^{2}}{16} -\frac{y^{2}}{4} =1$  , 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:

$\frac{(x-a)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1$

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.

$\sqrt{4^{2}+2^{2}}$ = $2\sqrt{5}$

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:

$(2\sqrt{5}, 0)$ . ∴ foci of $\frac{x^{2}}{16} -\frac{y^{2}}{4} =1$ is $(2\sqrt{5}, 0)$. correct option a.