Chapter 10.4, Problem 9AYU

Answer Problems 9-11 using the figure to the right.

The equation of the hyperbola is of the form

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

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

(c) $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

(d) $\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$

To determine

To find: The equation of the hyperbola from the given option:

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

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

c. $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

d. $\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$

Answer to Problem 9AYU

Solution:

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

Explanation of Solution

Given:

The hyperbola and the options.

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

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

c. $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

d. $\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$

Formula used:

Center	Transverse axis	Foci	Vertices	Equation
$\left(h,k\right)$	parallel to $y\text{-axis}$	$\left(h,k+c\right)$	$\left(h,k+a\right)$	$\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ $b²=c²-a²$
$\left(h,k\right)$	parallel to $y\text{-axis}$	$\left(h,k-c\right)$	$\left(h,k-a\right)$

Calculation:

From the given figure we see that, transverse axis is parallel to $y\text{-axis}$, hence the equation of the hyperbola is given by $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$.

Option $b$ is correct.