Question

write the equation of the hyperbola ( 36 x^{2}-4 y^{2}+216 x-32 y+116 = 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 x and y terms

Group the \(x\)-terms and \(y\)-terms together:
\(36x^2 + 216x - 4y^2 - 32y = -116\)

Step2: Factor out coefficients of squared terms

Factor out 36 from the \(x\)-terms and -4 from the \(y\)-terms:
\(36(x^2 + 6x) - 4(y^2 + 8y) = -116\)

Step3: Complete the square for x and y

For \(x\): \(x^2 + 6x\), take half of 6 (\(= 3\)), square it (\(= 9\)).
For \(y\): \(y^2 + 8y\), take half of 8 (\(= 4\)), square it (\(= 16\)).

Add and subtract the squared terms (remember to multiply by the factored coefficients):
\(36(x^2 + 6x + 9) - 4(y^2 + 8y + 16) = -116 + 36(9) - 4(16)\)

Step4: Simplify both sides

Calculate the right-hand side:
\(36(9) = 324\), \(-4(16) = -64\)
\(-116 + 324 - 64 = 144\)

Left-hand side becomes:
\(36(x + 3)^2 - 4(y + 4)^2 = 144\)

Step5: Divide by 144 to get standard form

Divide all terms by 144:
\(\frac{36(x + 3)^2}{144} - \frac{4(y + 4)^2}{144} = \frac{144}{144}\)
Simplify fractions:
\(\frac{(x + 3)^2}{4} - \frac{(y + 4)^2}{36} = 1\)

Answer:

\(h = -3\), \(k = -4\), \(a = 2\), \(b = 6\)

(Standard form: \(\boldsymbol{\frac{(x + 3)^2}{4} - \frac{(y + 4)^2}{36} = 1}\))