We are given that a vase is modeled by the following hyperbola:
[tex]\frac{x^{2}}{6.25}-\frac{\left(y-4\right)^{2}}{56.77}=1[/tex]
we are asked to determine the distance across the base. To do that we will first look at the graph of the equation:
Therefore, the base of the vase is the distance between the x-intercepts of the graph. To determine the x-intercepts we will set y = 0 in the equation. We get:
[tex]\frac{x^2}{6.25}-\frac{(0-4)^2}{56.77}=1[/tex]
Solving the operation on the parenthesis we get:
[tex]\frac{x^2}{6.25}-\frac{16}{56.77}=1[/tex]
Now we solve the fraction:
[tex]\frac{x^2}{6.25}-0.28=1[/tex]
Now we add 0.28 to both sides:
[tex]\begin{gathered} \frac{x^2}{6.25}=1+0.25 \\ \\ \frac{x^2}{6.25}=1.25 \end{gathered}[/tex]
Now we will multiply 6.25:
[tex]\begin{gathered} x^2=1.25(6.25) \\ x^2=7.81 \end{gathered}[/tex]
Taking square root to both sides:
[tex]\begin{gathered} x=\sqrt[]{7.81} \\ x=\pm2.8 \end{gathered}[/tex]
Therefore, the x-intercepts are -2.8 and 2.8.
Now we need to determine the distance between these two points. We will use the distance between two points in a line:
[tex]d=\lvert x_2-x_1\rvert[/tex]
Substituting the points we get:
[tex]d=\lvert2.8-(-2.8)\rvert=\lvert2.8+2.8\rvert=5.6[/tex]
Therefore, the distance is 5.6 inches and is related to the hyperbola in the sense that it is the distance between the x-intercepts.
