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Write the equation for the hyperbola with foci [tex]$(-12,6)$[/tex] and [tex]$(6,6)$[/tex] and vertices [tex][tex]$(-10,6)$[/tex][/tex] and [tex]$(4,6)$[/tex].

A. [tex]\frac{(x+3)^2}{7}-\frac{(y-6)^2}{5.66}=1[/tex]
B. [tex]\frac{(x+3)^2}{49}-\frac{(y-6)^2}{32}=1[/tex]
C. [tex]\frac{(y+3)^2}{49}-\frac{(x-6)^2}{32}=1[/tex]
D. [tex]\frac{(x+3)^2}{49}+\frac{(y-6)^2}{32}=1[/tex]


Sagot :

To write the equation for a hyperbola given its foci [tex]\((-12, 6)\)[/tex] and [tex]\((6, 6)\)[/tex] and its vertices [tex]\((-10, 6)\)[/tex] and [tex]\((4, 6)\)[/tex], we need to follow several steps. Let's go through them step-by-step.

### Step 1: Determine the center of the hyperbola

The center of the hyperbola is the midpoint between the foci.
- Let the coordinates of the foci be [tex]\(F_1 = (-12, 6)\)[/tex] and [tex]\(F_2 = (6, 6)\)[/tex].

[tex]\[ \text{Center} \left(h, k\right) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{-12 + 6}{2}, \frac{6 + 6}{2}\right) = (-3, 6) \][/tex]

### Step 2: Determine the distance [tex]\(2a\)[/tex] between the vertices

The distance between the vertices is [tex]\(2a\)[/tex].
- The coordinates of the vertices are [tex]\(V_1 = (-10, 6)\)[/tex] and [tex]\(V_2 = (4, 6)\)[/tex].

[tex]\[ 2a = \text{distance between the vertices} = |x_2 - x_1| = |4 - (-10)| = 14 \][/tex]

Thus, [tex]\(a = \frac{14}{2} = 7\)[/tex].

### Step 3: Determine the distance [tex]\(2c\)[/tex] between the foci

The distance between the foci is [tex]\(2c\)[/tex].
[tex]\[ 2c = \text{distance between the foci} = |x_2 - x_1| = |6 - (-12)| = 18 \][/tex]

Thus, [tex]\(c = \frac{18}{2} = 9\)[/tex].

### Step 4: Compute [tex]\(b\)[/tex] using the relationship [tex]\(c^2 = a^2 + b^2\)[/tex]

[tex]\[ c^2 = a^2 + b^2 \implies 9^2 = 7^2 + b^2 \implies 81 = 49 + b^2 \implies b^2 = 81 - 49 = 32 \implies b = \sqrt{32} = 4\sqrt{2} \][/tex]

### Step 5: Write the standard form of the equation of the hyperbola

Since the hyperbola opens along the x-axis (horizontal hyperbola):
[tex]\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \][/tex]

Substitute [tex]\(h = -3\)[/tex], [tex]\(k = 6\)[/tex], [tex]\(a = 7\)[/tex], [tex]\(b = \sqrt{32}\)[/tex]:
[tex]\[ \frac{(x + 3)^2}{7^2} - \frac{(y - 6)^2}{(\sqrt{32})^2} = 1 \][/tex]

Simplify:
[tex]\[ \frac{(x + 3)^2}{49} - \frac{(y - 6)^2}{32} = 1 \][/tex]

Therefore, the correct equation of the hyperbola is:
[tex]\[ \boxed{\frac{(x + 3)^2}{49} - \frac{(y - 6)^2}{32} = 1} \][/tex]