Question

1. transform the equation into the standard form for hyperbolas.\\( \frac { ( x - h ) ^ { 2 } } { a ^ { 2 } } - \frac { ( y - k ) ^ { 2 } } { b ^ { 2 } } = 1 \\).\\( 4 x ^ { 2 } - 3 y ^ { 2 } + 8 x + 6 y - 23 = 0 \\)\\( \frac { ( x - 1 ) ^ { 2 } } { 8 } - \frac { ( y - 1 ) ^ { 2 } } { 6 } = 1 \\)\\( \frac { ( x + 1 ) ^ { 2 } } { 6 } - \frac { ( y + 1 ) ^ { 2 } } { 8 } = 1 \\)\\( \frac { ( x + 1 ) ^ { 2 } } { 8 } - \frac { ( y + 1 ) ^ { 2 } } { 6 } = 1 \\)\\( \frac { ( x + 1 ) ^ { 2 } } { 6 } - \frac { ( y - 1 ) ^ { 2 } } { 8 } = 1 \\)

Explanation:

Step1: Group \(x\) and \(y\) terms

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Step2: Complete the square for \(x\) and \(y\)

For \(x\): \((x^{2}+2x)=(x + 1)^{2}-1\)
For \(y\): \((y^{2}-2y)=(y - 1)^{2}-1\)

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Step3: Divide by 24 to get the standard form

$$ \frac{4(x + 1)^{2}}{24}-\frac{3(y - 1)^{2}}{24}=1\Rightarrow\frac{(x + 1)^{2}}{6}-\frac{(y - 1)^{2}}{8}=1 $$

Answer:

\(\frac{(x + 1)^{2}}{6}-\frac{(y - 1)^{2}}{8}=1\) (the fourth option in the given choices)