Given the function of the conic section:
[tex]\mleft(y+2\mright)^2/16-\mleft(x-3\mright)^2/9=1[/tex]This conic section is a hyperbola.
Use this form below to determine the values used to find vertices and asymptotes of the hyperbola:
[tex]\frac{(x-h)^2}{a^2}\text{ - }\frac{(y-k)^2}{b^2}\text{ = }1[/tex]Match the values in this hyperbola to those of the standard form.
The variable h represents the x-offset from the origin b, k represents the y-offset from origin a.
We get,
a = 4
b = 3
k = 3
h = -2
A. The first focus of a hyperbola can be found by adding the distance of the center to a focus or c to h.
But first, let's determine the value of c. We will be using the formula below:
[tex]\sqrt[]{a^2+b^2}[/tex]Let's now determine the value of c.
[tex]\sqrt[]{a^2+b^2}\text{ = }\sqrt[]{4^2+3^2}\text{ = }\sqrt[]{16\text{ + 9}}\text{ = }\sqrt[]{25}[/tex][tex]\text{ c = 5}[/tex]Let's now determine the coordinates of the first foci:
[tex]\text{Coordinates of 1st Foci: (}h\text{ + c, k) = (-2 + 5, 3) = 3,3}[/tex]B. The second focus of a hyperbola can be found by subtracting c from h.
[tex]\text{ Coordinates of 2nd Foci: (h - c, k) = (-2 - 5, 3) = -7,3}[/tex]Therefore, the conic section has two focus and their coordinates are 3,3 and -7,3.
In other forms, the foci of the hyperbola is:
[tex]\text{ }(h\text{ }\pm\text{ }\sqrt[]{a^2+b^2},\text{ k) or (-2 }\pm\text{ 5, 3)}[/tex]Therefore, the answer is letter B.
