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The hyperbola [tex]\frac{(x-5)^2}{7} - \frac{(y+3)^2}{9} = 1[/tex] is shifted to the right by 4 units and upward by 3 units.

The new center of the hyperbola is ( [tex]$\qquad$[/tex] , [tex]$\qquad$[/tex] ).


Sagot :

To solve the problem of finding the new center of the hyperbola after it has been shifted, follow these steps:

1. Identify the original center:
The given equation of the hyperbola is [tex]\(\frac{(x-5)^2}{7} - \frac{(y+3)^2}{9} = 1\)[/tex]. This equation is in standard form for a hyperbola centered at [tex]\((h, k)\)[/tex].
Comparing it with the standard form [tex]\(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)[/tex], we identify:
- [tex]\(h = 5\)[/tex]
- [tex]\(k = -3\)[/tex]

Therefore, the original center of the hyperbola is [tex]\((5, -3)\)[/tex].

2. Apply the horizontal shift:
The hyperbola is shifted 4 units to the right. To find the new [tex]\(x\)[/tex]-coordinate of the center:
- Original [tex]\(x\)[/tex]-coordinate: 5
- Shift to the right: +4

Therefore, the new [tex]\(x\)[/tex]-coordinate is [tex]\(5 + 4 = 9\)[/tex].

3. Apply the vertical shift:
The hyperbola is shifted 3 units upward. To find the new [tex]\(y\)[/tex]-coordinate of the center:
- Original [tex]\(y\)[/tex]-coordinate: -3
- Shift upward: +3

Therefore, the new [tex]\(y\)[/tex]-coordinate is [tex]\(-3 + 3 = 0\)[/tex].

Combining these two results, the new center of the hyperbola is [tex]\((9, 0)\)[/tex].

Hence, the new center of the hyperbola is [tex]\(( \ 9 \ , \ 0 \ )\)[/tex].
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