The standard form of a quadratic equation is: \[ y = ax^2 + bx + c \] where \( a = 2 \), \( b = -12 \), and \( c = 26 \). The formula for the vertex is: \[ x = -\frac{b}{2a} \] \[ x = -\frac{-12}{2(2)} = \frac{12}{4} = 3 \] Now, substituting \( x = 3 \) in the given equation: \[ y = 2(3)^2 - 12(3) + 26 \] \[ y = 2(9) - 36 + 26 = 18 - 36 + 26 = 8 \] Thus, the vertex is \( (3,8) \).
Final Answer: \[ \boxed{(3,8)} \]
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to: