Question

question 18 · 1 point
find the equation of the hyperbola with vertices (2,4) and (2, - 8) and foci (2,6) and (2, - 10).
select the correct answer below:
\\(\frac{(x + 2)^2}{36}-\frac{(y + 2)^2}{28}=1\\)
\\(\frac{(x - 2)^2}{36}-\frac{(y + 2)^2}{28}=1\\)
\\(\frac{(y + 2)^2}{36}-\frac{(x - 2)^2}{28}=1\\)
\\(\frac{(y + 2)^2}{28}-\frac{(x + 2)^2}{36}=1\\)
\\(\frac{(y - 2)^2}{28}-\frac{(x + 2)^2}{36}=1\\)
\\(\frac{(x + 2)^2}{28}-\frac{(y - 2)^2}{36}=1\\)

Explanation:

Step1: Determine the center

The vertices are $(2,4)$ and $(2,-8)$. The center of the hyperbola is the mid - point of the vertices. The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Here $x_1=x_2 = 2$, and $\frac{4+( - 8)}{2}=\frac{4 - 8}{2}=-2$. So the center is $(2,-2)$.

Step2: Find the value of $a$

The distance between the center $(2,-2)$ and a vertex $(2,4)$ (or $(2,-8)$) is $a$. Using the distance formula for points with the same $x$ - coordinate $d=\vert y_2 - y_1\vert$. Here $a=\vert4-( - 2)\vert=\vert4 + 2\vert = 6$, so $a^{2}=36$.

Step3: Find the value of $c$

The foci are $(2,6)$ and $(2,-10)$. The distance between the center $(2,-2)$ and a focus $(2,6)$ (or $(2,-10)$) is $c$. Using the distance formula for points with the same $x$ - coordinate, $c=\vert6-( - 2)\vert=\vert6 + 2\vert = 8$.

Step4: Calculate $b^{2}$

We know the relationship $c^{2}=a^{2}+b^{2}$ for a hyperbola. Substituting $a^{2}=36$ and $c^{2}=64$ into the formula, we get $b^{2}=c^{2}-a^{2}=64 - 36=28$.

Step5: Determine the equation

Since the vertices and foci have the same $x$ - coordinate, the hyperbola has a vertical transverse axis. The standard form of a hyperbola with a vertical transverse axis and center $(h,k)$ is $\frac{(y - k)^2}{a^{2}}-\frac{(x - h)^2}{b^{2}}=1$. Here $h = 2,k=-2,a^{2}=36,b^{2}=28$, so the equation is $\frac{(y + 2)^2}{36}-\frac{(x - 2)^2}{28}=1$.

Answer:

C. $\frac{(y + 2)^2}{36}-\frac{(x - 2)^2}{28}=1$