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The equation [tex]\frac{x^2}{24^2} - \frac{y^2}{[-]^2} = 1[/tex] represents a hyperbola centered at the origin with a directrix of [tex]x = \frac{575}{25}[/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.



Answer :

GinnyAnswer
To solve this problem, we need to find the value of [tex]\( b \)[/tex] to correctly complete the equation for the hyperbola. Let's walk through the steps one by one:

1. Given Information:
- The standard form of the hyperbola centered at the origin:
[tex]\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \][/tex]
- The value of [tex]\( a \)[/tex] is given as [tex]\( 24 \)[/tex].
- The directrix is given as [tex]\( x = \frac{575}{25} = 23 \)[/tex].

2. Directrix Relationship:
- For a hyperbola, the directrix [tex]\( x = \frac{a^2}{c} \)[/tex] can be used to find [tex]\( c \)[/tex].

Using the given directrix:
[tex]\[ x = \frac{a^2}{c} \quad \Rightarrow \quad 23 = \frac{24^2}{c} \][/tex]
Solving for [tex]\( c \)[/tex]:
[tex]\[ c = \frac{576}{23} \approx 25.043 \][/tex]

3. Relationship Between [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
- For a hyperbola, the relationship is [tex]\( c^2 = a^2 + b^2 \)[/tex].
- We already have [tex]\( a = 24 \)[/tex] and [tex]\( c \approx 25.043 \)[/tex].

Finding [tex]\( b^2 \)[/tex]:
[tex]\[ c^2 = a^2 + b^2 \quad \Rightarrow \quad (25.043)^2 = 24^2 + b^2 \][/tex]
[tex]\[ 627.173 = 576 + b^2 \][/tex]
Solving for [tex]\( b^2 \)[/tex]:
[tex]\[ b^2 = 627.173 - 576 \approx 51.174 \][/tex]
Thus:
[tex]\[ b \approx \sqrt{51.174} \approx 7.154 \][/tex]

4. Completing the Equation:
- The hyperbola has the form [tex]\( \frac{x^2}{24^2} - \frac{y^2}{b^2} = 1 \)[/tex].
- The value of [tex]\( b \approx 7.154 \)[/tex] completes the blank space.

So, the complete equation is:
[tex]\[ \frac{x^2}{24^2} - \frac{y^2}{7.154^2} = 1 \][/tex]

To summarize, the positive value [tex]\( \boxed{7.154} \)[/tex] correctly fills in the blank in the hyperbola's equation.

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