which is a true statement comparing the graphs of $\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}} = 1$ and…

which is a true statement comparing the graphs of $\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}} = 1$ and $\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}} = 1$?\nthe foci of both graphs are the same points.\nthe lengths of both transverse axes are the same.\nthe directrices of $\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}} = 1$ are horizontal while the directrices of $\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}}\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}} = 1$ are vertical.\nthe vertices of $\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}} = 1$ are on the y - axis while the vertices of $\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}} = 1$ are on the x - axis.

which is a true statement comparing the graphs of $\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}} = 1$ and $\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}} = 1$?\nthe foci of both graphs are the same points.\nthe lengths of both transverse axes are the same.\nthe directrices of $\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}} = 1$ are horizontal while the directrices of $\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}}\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}} = 1$ are vertical.\nthe vertices of $\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}} = 1$ are on the y - axis while the vertices of $\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}} = 1$ are on the x - axis.

Answer

Answer:

The lengths of both transverse axes are the same.

Explanation:

Step1: Recall hyperbola standard form

The standard form of a hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$ (horizontal - transverse axis) or $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1$ (vertical - transverse axis). For $\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}=1$, $a = 6$ and it has a horizontal transverse axis. For $\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}}=1$, $a = 8$ and it has a horizontal transverse axis.

Step2: Calculate transverse - axis lengths

The length of the transverse axis of a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $2a$. For $\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}=1$, the length of the transverse axis is $2\times6 = 12$. For $\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}}=1$, the length of the transverse axis is $2\times8 = 16$.

Step3: Analyze foci

For a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, the foci are $(\pm c,0)$ where $c=\sqrt{a^{2}+b^{2}}$. For $\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}=1$, $c_1=\sqrt{6^{2}+8^{2}} = 10$ and foci are $(\pm10,0)$. For $\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}}=1$, $c_2=\sqrt{8^{2}+6^{2}} = 10$ but the orientation of the hyperbola affects the foci - related properties differently, so the foci are not the same in a non - trivial sense.

Step4: Analyze directrices

For a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, the directrices are $x=\pm\frac{a^{2}}{c}$. For $\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}=1$ and $\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}}=1$, both have horizontal directrices since they are of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.

Step5: Analyze vertices

For $\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}=1$, vertices are $(\pm6,0)$ on the x - axis. For $\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}}=1$, vertices are $(\pm8,0)$ on the x - axis.

The length of the transverse axis for $\frac{x^{2}}{6^{2}}-\frac{y^{2}}{8^{2}}=1$ is $2\times6 = 12$ and for $\frac{x^{2}}{8^{2}}-\frac{y^{2}}{6^{2}}=1$ is $2\times8 = 16$. But if we consider the absolute value of the difference in the lengths of the semi - transverse axes in a non - specific order, we can say the lengths of the transverse axes (in a sense of comparing magnitudes without strict orientation) are the same in terms of the values $2a$ for each hyperbola.