Question

write the equation of the hyperbola ( 16 x^{2}-4 y^{2}-96 x + 8 y + 76 = 0 ) in standard form
\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}} = 1,
where:
( h = )
( k = )
( a = )
( b = )
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Explanation:

Step1: Group the x and y terms

$$16x^{2}-96x - 4y^{2}+8y=-76$$

Step2: Factor out the coefficients of \(x^{2}\) and \(y^{2}\)

$$16(x^{2}-6x)-4(y^{2}-2y)=-76$$

Step3: Complete the square for the x and y expressions

For \(x^{2}-6x\), \((\frac{-6}{2})^{2}=9\). For \(y^{2}-2y\), \((\frac{-2}{2})^{2}=1\)
$$16(x^{2}-6x + 9)-4(y^{2}-2y+1)=-76+16\times9-4\times1$$

Step4: Simplify the right - hand side

$$16(x - 3)^{2}-4(y - 1)^{2}=64$$

Step5: Divide through by 64 to get the standard form

$$\frac{(x - 3)^{2}}{4}-\frac{(y - 1)^{2}}{16}=1$$

Answer:

\(h = 3\), \(k = 1\), \(a = 2\), \(b = 4\)