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The equation [tex]\frac{x^2}{24^2} - \frac{y^2}{[\ldots]^2} = 1[/tex] represents a hyperbola centered at the origin with a directrix of [tex]x = \frac{576}{26}[/tex].

\begin{tabular}{|l|l|}
\hline
Vertices: [tex](-a, 0), (a, 0)[/tex] & Vertices: [tex](0, -a), (0, a)[/tex] \\
Foci: [tex](-c, 0), (c, 0)[/tex] & Foci: [tex](0, -c), (0, c)[/tex] \\
Asymptotes: [tex]y = \pm \frac{b}{a} x[/tex] & Asymptotes: [tex]y = \pm \frac{a}{b} x[/tex] \\
Directrices: [tex]x = \pm \frac{a^2}{c}[/tex] & Directrices: [tex]y = \pm \frac{a^2}{c}[/tex] \\
Standard Equation: [tex]\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1[/tex] & Standard Equation: [tex]\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1[/tex] \\
\hline
\end{tabular}

The positive value [tex]\square[/tex] correctly fills in the blank in the equation.

Sagot :

GinnyAnswer
To find the value that correctly fills in the blank in the given hyperbola equation \(\frac{x^2}{24^2} - \frac{y^2}{[\ldots]^2} = 1\), we will go through the necessary steps to determine the missing value step-by-step.

1. Identify the given values:
- The term \(\frac{x^2}{24^2}\) tells us that \(a^2 = 24^2\), so \(a = 24\).
- The directrix is given as \(x = \frac{576}{26}\).

2. Determine \(c\):
- For a hyperbola with the equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), and directrix \(x = \frac{a^2}{c}\), we can set up the equation for the directrix:
[tex]\[ x = \frac{a^2}{c} \][/tex]
- We know \(a^2 = 576\), so:
[tex]\[ \frac{576}{c} = \frac{576}{26} \][/tex]
- Solving for \(c\):
[tex]\[ c = 26 \][/tex]

3. Find \(b^2\):
- For hyperbolas, the relationship between \(a\), \(b\), and \(c\) is given by:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
- Substitute the known values \(a = 24\) and \(c = 26\):
[tex]\[ 26^2 = 24^2 + b^2 \][/tex]
- Calculate \(26^2\) and \(24^2\):
[tex]\[ 676 = 576 + b^2 \][/tex]
- Solve for \(b^2\):
[tex]\[ b^2 = 676 - 576 = 100 \][/tex]

4. State the complete equation:
- Now that we have \(b^2 = 100\), we can fill in the blank in the original equation:
[tex]\[ \frac{x^2}{24^2} - \frac{y^2}{100} = 1 \][/tex]

Thus, the positive value that correctly fills in the blank in the equation [tex]\(\frac{x^2}{24^2} - \frac{y^2}{[\ldots]^2} = 1\)[/tex] is [tex]\(10\)[/tex] (since [tex]\(10^2 = 100\)[/tex]).
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