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The equation [tex]\frac{x^2}{24^2}-\frac{y^2}{[\square]^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 that correctly fills in the blank in the equation is [tex]\square[/tex].



Answer :

GinnyAnswer
To solve for the positive value that correctly fills in the blank in the equation [tex]\(\frac{x^2}{24^2} - \frac{y^2}{[\square]^2} = 1\)[/tex], follow these steps:

1. Identify the given parameters:
- The hyperbola equation is [tex]\(\frac{x^2}{24^2} - \frac{y^2}{[\square]^2} = 1\)[/tex], which indicates [tex]\(a = 24\)[/tex] and we need to find [tex]\(b\)[/tex].
- The directrix is given by the equation [tex]\(x = \frac{576}{26}\)[/tex].

2. Compute the directrix value:
- Calculate the directrix value: [tex]\(x = \frac{576}{26}\)[/tex]:
[tex]\[ x = \frac{576}{26} = 22.153846153846153 \][/tex]

3. Compute the value of [tex]\(c\)[/tex]:
- The relationship between the directrix and the hyperbola parameters is given by the formula [tex]\(x = \frac{a^2}{c}\)[/tex], where [tex]\(c\)[/tex] is the distance from the center to the foci:
[tex]\[ 22.153846153846153 = \frac{24^2}{c} \][/tex]
- Solve for [tex]\(c\)[/tex]:
[tex]\[ c = \frac{24^2}{22.153846153846153} \][/tex]
- The value obtained for [tex]\(c\)[/tex] is:
[tex]\[ c = 26.0 \][/tex]

4. Use the relationship between [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
- In a hyperbola, the relationship between [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] is given by:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
- Substitute [tex]\(a = 24\)[/tex] and [tex]\(c = 26.0\)[/tex]:
[tex]\[ 26^2 = 24^2 + b^2 \][/tex]
- Simplify and solve for [tex]\(b^2\)[/tex]:
[tex]\[ 676 = 576 + b^2 \][/tex]
[tex]\[ b^2 = 676 - 576 \][/tex]
[tex]\[ b^2 = 100 \][/tex]

5. Find [tex]\(b\)[/tex]:
- Calculate [tex]\(b\)[/tex]:
[tex]\[ b = \sqrt{100} = 10.0 \][/tex]

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

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