foci of [tex]\frac{x^{2}}{16} -\frac{y^{2}}{4} =1[/tex] is [tex](2\sqrt{5}, 0)[/tex]. correct option a.
Step-by-step explanation:
Complete equation of Hyperbola for the above question is x2/16-y2/4 = 1 or [tex]\frac{x^{2}}{16} -\frac{y^{2}}{4} =1[/tex] , 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:
[tex]\frac{(x-a)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1[/tex]
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:
[tex]\sqrt{a^{2}+b^{2}}[/tex]
Substitute the value of a and b in the formula.
[tex]\sqrt{4^{2}+2^{2}}[/tex] = [tex]2\sqrt{5}[/tex]
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:
[tex](2\sqrt{5}, 0)[/tex] . ∴ foci of [tex]\frac{x^{2}}{16} -\frac{y^{2}}{4} =1[/tex] is [tex](2\sqrt{5}, 0)[/tex]. correct option a.
