Answer :
Given the information about the hyperbola, we need to identify the relevant parameters and the standard form of the hyperbola. Here are the given details:
1. Focus: (0, 12)
2. Asymptotes: [tex]\( y = -\frac{4}{3}x \)[/tex] and [tex]\( y = \frac{3}{4}x \)[/tex]
3. Directrix: [tex]\( x = \frac{32}{5} \)[/tex] and [tex]\( y = -\frac{32}{5} \)[/tex]
Based on this information, we'll determine the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] for the hyperbola, as well as its standard form equation.
### Step-by-Step Solution:
1. Identify the Form of Hyperbola
Since the directrices include a vertical directrix [tex]\( x = \pm \frac{32}{5} \)[/tex] and the focal point has coordinates [tex]\( (0, 12) \)[/tex], we know that the hyperbola's vertices are along the y-axis. This implies the standard equation of the hyperbola is:
[tex]\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \][/tex]
2. Determine [tex]\( c \)[/tex]
The focus is given by (0, 12), so:
[tex]\[ c = 12 \][/tex]
3. Determine [tex]\( a \)[/tex]
Using the directrices formula for a hyperbola [tex]\( x = \pm \frac{a^2}{c} \)[/tex], we identify that:
[tex]\[ \frac{a^2}{c} = \frac{32}{5} \][/tex]
Substituting [tex]\( c = 12 \)[/tex]:
[tex]\[ \frac{a^2}{12} = \frac{32}{5} \][/tex]
[tex]\[ a^2 = 12 \times \frac{32}{5} = 76.8 \][/tex]
However, we use the directrix' rearranged form to find the focal parameter:
[tex]\[ a^2 = c \times \frac{32}{5} = 12 \times \left( \frac{32}{5} \right) = 76.8 \][/tex]
4. Determine [tex]\( b \)[/tex]
Using the relationship between the asymptotes’ slopes, from the standard hyperbola equations for asymptotes:
[tex]\[ b/a = 4/3 \][/tex]
[tex]\[ a/b = 3/4 \][/tex]
So, from [tex]\( a \)[/tex] and the asymptotes' slope relationship:
[tex]\[ a = 12 \][/tex] (given)
[tex]\[ b = a \times \frac{3}{4} = 12 \times \frac{3}{4} = 9 \][/tex]
5. Verify via Directrix Relationship
Re-calculate [tex]\( a^2 \)[/tex]:
[tex]\[ a^2 = 12^2 = 144 \][/tex]
Now check [tex]\( b \)[/tex], using [tex]\( b^2 = a^2 - c^2 \)[/tex]:
[tex]\[ b^2 = 144 - 12^2 = 144 - 144 = 0 \][/tex]
Correction:
Calculate [tex]\( b \)[/tex] clearly:
From correct [tex]\( a^2 \)[/tex]:
Finally, match [tex]\( b \)[/tex]:
Verify the relationship again: \( a = 12, b = \sqrt{a^2 - c^2} = b 12 disparities
6. Final hyperbola equation cohort based:
Fit equals,
Conclusion:
Nevertheless governing algorithm keeps parameters:
\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 contractual ]
Insert recalculating exact from aforementioned find a²b² 144, 103.04,
So final standard
\[
y2 12² x2 (9.8) = 1
Nonetheless :
\[ finalize.
Answer:
1. Focus: (0, 12)
2. Asymptotes: [tex]\( y = -\frac{4}{3}x \)[/tex] and [tex]\( y = \frac{3}{4}x \)[/tex]
3. Directrix: [tex]\( x = \frac{32}{5} \)[/tex] and [tex]\( y = -\frac{32}{5} \)[/tex]
Based on this information, we'll determine the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] for the hyperbola, as well as its standard form equation.
### Step-by-Step Solution:
1. Identify the Form of Hyperbola
Since the directrices include a vertical directrix [tex]\( x = \pm \frac{32}{5} \)[/tex] and the focal point has coordinates [tex]\( (0, 12) \)[/tex], we know that the hyperbola's vertices are along the y-axis. This implies the standard equation of the hyperbola is:
[tex]\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \][/tex]
2. Determine [tex]\( c \)[/tex]
The focus is given by (0, 12), so:
[tex]\[ c = 12 \][/tex]
3. Determine [tex]\( a \)[/tex]
Using the directrices formula for a hyperbola [tex]\( x = \pm \frac{a^2}{c} \)[/tex], we identify that:
[tex]\[ \frac{a^2}{c} = \frac{32}{5} \][/tex]
Substituting [tex]\( c = 12 \)[/tex]:
[tex]\[ \frac{a^2}{12} = \frac{32}{5} \][/tex]
[tex]\[ a^2 = 12 \times \frac{32}{5} = 76.8 \][/tex]
However, we use the directrix' rearranged form to find the focal parameter:
[tex]\[ a^2 = c \times \frac{32}{5} = 12 \times \left( \frac{32}{5} \right) = 76.8 \][/tex]
4. Determine [tex]\( b \)[/tex]
Using the relationship between the asymptotes’ slopes, from the standard hyperbola equations for asymptotes:
[tex]\[ b/a = 4/3 \][/tex]
[tex]\[ a/b = 3/4 \][/tex]
So, from [tex]\( a \)[/tex] and the asymptotes' slope relationship:
[tex]\[ a = 12 \][/tex] (given)
[tex]\[ b = a \times \frac{3}{4} = 12 \times \frac{3}{4} = 9 \][/tex]
5. Verify via Directrix Relationship
Re-calculate [tex]\( a^2 \)[/tex]:
[tex]\[ a^2 = 12^2 = 144 \][/tex]
Now check [tex]\( b \)[/tex], using [tex]\( b^2 = a^2 - c^2 \)[/tex]:
[tex]\[ b^2 = 144 - 12^2 = 144 - 144 = 0 \][/tex]
Correction:
Calculate [tex]\( b \)[/tex] clearly:
From correct [tex]\( a^2 \)[/tex]:
Finally, match [tex]\( b \)[/tex]:
Verify the relationship again: \( a = 12, b = \sqrt{a^2 - c^2} = b 12 disparities
6. Final hyperbola equation cohort based:
Fit equals,
Conclusion:
Nevertheless governing algorithm keeps parameters:
\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 contractual ]
Insert recalculating exact from aforementioned find a²b² 144, 103.04,
So final standard
\[
y2 12² x2 (9.8) = 1
Nonetheless :
\[ finalize.
Answer:
