Isolate y from the equation:
[tex]\frac{(x-2)^2}{9}-\frac{(y-1)^2}{4}=1[/tex][tex]\Rightarrow-\frac{(y-1)^2}{4}=1-\frac{(x-2)^2}{9}[/tex][tex]\begin{gathered} \Rightarrow(y-1)^2=(1-\frac{(x-2)^2}{9})(-4) \\ =\frac{4}{9}(x-2)^2-4 \end{gathered}[/tex][tex]\begin{gathered} \Rightarrow y-1=\pm\sqrt[]{\frac{4}{9}(x-2)^2-4} \\ \Rightarrow y=1\pm\sqrt[]{\frac{4}{9}(x-2)^2-4} \end{gathered}[/tex]
Substitute x=10 to find the two possible values for y:
[tex]\begin{gathered} y=1\pm\sqrt[]{\frac{4}{9}(10-2)^2-4} \\ =1\pm\sqrt[]{\frac{4}{9}(8)^2-4} \\ =1\pm\sqrt[]{\frac{4}{9}(64)^{}-4} \\ =1\pm\sqrt[]{\frac{256}{9}-4} \\ =1\pm\sqrt[]{\frac{256-36}{9}} \\ =1\pm\sqrt[]{\frac{220}{9}} \\ =1\pm\frac{2\sqrt[]{55}}{3} \end{gathered}[/tex]
The aproximate values of y are:
[tex]\begin{gathered} y_1\approx5.944\ldots \\ y_2\approx-3.944\ldots \end{gathered}[/tex]
To the nearest whole number, we can see that the second value of y is approximately -4.
