Question

find the center and vertices for this hyperbola:
4y² - 49x² - 392x + 16y - 964 = 0
center = ( ?, )
vertices = ( , ) and ( , )

Explanation:

Step1: Group x and y terms

Group the \(x\)-terms and \(y\)-terms together:
\(-49x^2 - 392x + 4y^2 + 16y = 964\)
Factor out the coefficients of the squared terms from each group:
\(-49(x^2 + 8x) + 4(y^2 + 4y) = 964\)

Step2: Complete the square for x and y

For the \(x\)-group: \(x^2 + 8x\), take half of 8 (\(4\)), square it (\(16\)).
For the \(y\)-group: \(y^2 + 4y\), take half of 4 (\(2\)), square it (\(4\)).

Add and subtract the squared terms (note the factor outside each group!):
\(-49(x^2 + 8x + 16 - 16) + 4(y^2 + 4y + 4 - 4) = 964\)
Simplify by distributing the factors:
\(-49[(x + 4)^2 - 16] + 4[(y + 2)^2 - 4] = 964\)
\(-49(x + 4)^2 + 784 + 4(y + 2)^2 - 16 = 964\)
Combine constants: \(784 - 16 = 768\), so:
\(-49(x + 4)^2 + 4(y + 2)^2 + 768 = 964\)
Subtract 768 from both sides:
\(-49(x + 4)^2 + 4(y + 2)^2 = 196\)

Step3: Rewrite in standard hyperbola form

Divide both sides by 196 to get 1 on the right:
\(\frac{4(y + 2)^2}{196} - \frac{49(x + 4)^2}{196} = 1\)
Simplify fractions:
\(\frac{(y + 2)^2}{49} - \frac{(x + 4)^2}{4} = 1\)

This is a vertical hyperbola (since the \(y\)-term is positive) with standard form \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\), where center \((h, k) = (-4, -2)\), \(a^2 = 49\) (so \(a = 7\)), \(b^2 = 4\) (so \(b = 2\)).

Step4: Find the center and vertices

• Center: From \((h, k)\) in the standard form, \(h = -4\), \(k = -2\), so center is \((-4, -2)\).
• Vertices: For a vertical hyperbola, vertices are \((h, k \pm a)\). Substitute \(h = -4\), \(k = -2\), \(a = 7\):

\(k + a = -2 + 7 = 5\), \(k - a = -2 - 7 = -9\).
Thus, vertices are \((-4, 5)\) and \((-4, -9)\).

Answer:

Center = \((-4, -2)\)
Vertices = \((-4, 5)\) and \((-4, -9)\)