Question

1. use the hyperbola equation $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$ to find the $y$-values to the nearest integer from the given $x$ and $y$-values in the table. $\frac{(x - 2)^2}{9}-\frac{(y - 1)^2}{4}=1$ $\

$$\begin{array}{cc}x&y\\10&6\\10&\\_\\end{array}$$

$ $-4$ $2$ $-6$ $4$

Explanation:

Step1: Substitute \(x = 10\) into the hyperbola equation

Given \(\frac{(x - 2)^2}{9}-\frac{(y - 1)^2}{4}=1\), when \(x = 10\), we have \(\frac{(10 - 2)^2}{9}-\frac{(y - 1)^2}{4}=1\).
First, calculate \(\frac{(10 - 2)^2}{9}=\frac{64}{9}\).
The equation becomes \(\frac{64}{9}-\frac{(y - 1)^2}{4}=1\).

Step2: Solve for \((y - 1)^2\)

Rearrange the equation: \(\frac{(y - 1)^2}{4}=\frac{64}{9}-1=\frac{64 - 9}{9}=\frac{55}{9}\).
Then \((y - 1)^2=\frac{55\times4}{9}=\frac{220}{9}\approx24.44\).

Step3: Solve for \(y\)

Take the square - root: \(y - 1=\pm\sqrt{\frac{220}{9}}\approx\pm4.94\).
Case 1: \(y-1\approx4.94\), then \(y\approx4.94 + 1=5.94\approx6\) (already given).
Case 2: \(y - 1\approx - 4.94\), then \(y\approx-4.94+1=-3.94\approx - 4\).

Answer:

A. \(-4\)