Question

What are the vertices of the hyperbola given by the equation 16x2 - y2 + 96x + 10y + 103 = 0?

Answer 1

16x² – y² + 96x + 10y + 103 = 0
You need to put the equation into standard hyperbola form.

regroup terms
16x² + 96x – y² + 10y = -103

factor out leading coefficients
16(x² + 6x) – (y² - 10y) = -103

complete the squares
16(x² + 6x + 3²) – (y² - 10y + (-5)²) = -103 +16(3²) - (-5)²
16(x + 3)² – (y - 5)² = 16

divide by 16 to set righthand-side to 1
16(x + 3)²/16 – (y - 5)²/16 = 16/16
(x + 3)² – (y - 5)²/16 = 1

express denominators as squares
(x + 3)²/1² – (y - 5)²/4² = 1
center at (-3, 5)
The x term is positive, so:
the hyperbola is horizontal.
vertices (-3±1, 5)
co-vertices (-3, 5±4)