Question

The centre of the hyperbola \(16x^2 - 4y^2 + 64x - 24y - 36 = 0\) is at the point:

• A. \((-2, -3)\)
• B. \((-4, -6)\)
• C. \( (2, 3) \)
• D. \( (4, 6) \)
• E. \( (2, 6) \)

Hint:

To find the center of a hyperbola, always complete the square for both \(x\) and \(y\) components. Remember, the form \((x-h)^2 - \frac{(y-k)^2}{a^2}\) or \(\frac{(y-k)^2}{a^2} - (x-h)^2\) reveals the center \((h, k)\).

Answer 1

The Correct Option is A

To find the center of the hyperbola, complete the square for the \(x\) and \(y\) terms in the equation: \[ 16x^2 + 64x - 4y^2 - 24y - 36 = 0 \] Group and complete the square: \[ 16(x^2 + 4x) - 4(y^2 + 6y) - 36 = 0 \] Complete the square inside the parentheses: \[ 16((x+2)^2 - 4) - 4((y+3)^2 - 9) - 36 = 0 \] Simplify: \[ 16(x+2)^2 - 64 - 4(y+3)^2 + 36 - 36 = 0 \] \[ 16(x+2)^2 - 4(y+3)^2 - 64 = 0 \] Further simplify to get the standard form: \[ 16(x+2)^2 - 4(y+3)^2 = 64 \] \[ (x+2)^2 - \frac{(y+3)^2}{4} = 4 \] The center of the hyperbola in the standard form \((x-h)^2 - \frac{(y-k)^2}{a^2} = 1\) or \(\frac{(y-k)^2}{a^2} - (x-h)^2 = 1\) is \((h, k)\). 
Here, it translates to \((-2, -3)\).