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Answer :
To find the standard equation of the hyperbola, let's start by working through the given equation:
The equation given is:
[tex]\[ 25y^2 + 100x - 100y - 625 = 25x^2 \][/tex]
First, rearrange it to bring all terms to one side:
[tex]\[ 25y^2 - 25x^2 + 100x - 100y - 625 = 0 \][/tex]
Reorganize the terms to group the y terms and the x terms:
[tex]\[ 25(y^2 - 4y) - 25(x^2 - 4x) = 625 \][/tex]
We need to complete the square for both the [tex]\( y \)[/tex] terms and the [tex]\( x \)[/tex] terms.
1. Completing the square for the y terms ([tex]\( y^2 - 4y \)[/tex]):
- Take the coefficient of [tex]\( y \)[/tex] (which is -4), divide it by 2, and square the result. [tex]\((-4/2)^2 = 4\)[/tex].
- Add and subtract 4 inside the bracket:
[tex]\[ y^2 - 4y = (y-2)^2 - 4 \][/tex]
2. Completing the square for the x terms ([tex]\( x^2 - 4x \)[/tex]):
- Take the coefficient of [tex]\( x \)[/tex] (which is -4), divide it by 2, and square the result. [tex]\((-4/2)^2 = 4\)[/tex].
- Add and subtract 4 inside the bracket:
[tex]\[ x^2 - 4x = (x-2)^2 - 4 \][/tex]
Substitute these completed squares back into the equation:
[tex]\[ 25((y-2)^2 - 4) - 25((x-2)^2 - 4) = 625 \][/tex]
Now expand and simplify:
[tex]\[ 25(y-2)^2 - 100 - 25(x-2)^2 + 100 = 625 \][/tex]
These constant terms (-100 and +100) cancel each other out:
[tex]\[ 25(y-2)^2 - 25(x-2)^2 = 625 \][/tex]
To put this into the standard form of a hyperbola [tex]\((\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1)\)[/tex], divide the entire equation by 625:
[tex]\[ \frac{(y-2)^2}{25} - \frac{(x-2)^2}{25} = 1 \][/tex]
Thus, the standard equation of the hyperbola is:
[tex]\[ \frac{(y-2)^2}{25} - \frac{(x-2)^2}{25} = 1 \][/tex]
This represents a hyperbola centered at [tex]\((2, 2)\)[/tex] with both the vertical and horizontal distances corresponding to [tex]\(\sqrt{25} = 5\)[/tex].
The equation given is:
[tex]\[ 25y^2 + 100x - 100y - 625 = 25x^2 \][/tex]
First, rearrange it to bring all terms to one side:
[tex]\[ 25y^2 - 25x^2 + 100x - 100y - 625 = 0 \][/tex]
Reorganize the terms to group the y terms and the x terms:
[tex]\[ 25(y^2 - 4y) - 25(x^2 - 4x) = 625 \][/tex]
We need to complete the square for both the [tex]\( y \)[/tex] terms and the [tex]\( x \)[/tex] terms.
1. Completing the square for the y terms ([tex]\( y^2 - 4y \)[/tex]):
- Take the coefficient of [tex]\( y \)[/tex] (which is -4), divide it by 2, and square the result. [tex]\((-4/2)^2 = 4\)[/tex].
- Add and subtract 4 inside the bracket:
[tex]\[ y^2 - 4y = (y-2)^2 - 4 \][/tex]
2. Completing the square for the x terms ([tex]\( x^2 - 4x \)[/tex]):
- Take the coefficient of [tex]\( x \)[/tex] (which is -4), divide it by 2, and square the result. [tex]\((-4/2)^2 = 4\)[/tex].
- Add and subtract 4 inside the bracket:
[tex]\[ x^2 - 4x = (x-2)^2 - 4 \][/tex]
Substitute these completed squares back into the equation:
[tex]\[ 25((y-2)^2 - 4) - 25((x-2)^2 - 4) = 625 \][/tex]
Now expand and simplify:
[tex]\[ 25(y-2)^2 - 100 - 25(x-2)^2 + 100 = 625 \][/tex]
These constant terms (-100 and +100) cancel each other out:
[tex]\[ 25(y-2)^2 - 25(x-2)^2 = 625 \][/tex]
To put this into the standard form of a hyperbola [tex]\((\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1)\)[/tex], divide the entire equation by 625:
[tex]\[ \frac{(y-2)^2}{25} - \frac{(x-2)^2}{25} = 1 \][/tex]
Thus, the standard equation of the hyperbola is:
[tex]\[ \frac{(y-2)^2}{25} - \frac{(x-2)^2}{25} = 1 \][/tex]
This represents a hyperbola centered at [tex]\((2, 2)\)[/tex] with both the vertical and horizontal distances corresponding to [tex]\(\sqrt{25} = 5\)[/tex].
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