Find the vertex and directrix of the parabola. $x ^ { 2 } - 18 x - 12 y + 33 = 0$
A)vertex: $( - 9,4 )$ directrix: y=1
B)vertex: $( - 9,4 )$ directrix: y=12
C)vertex: $( - 9,4 )$ directrix: y=6
D)vertex: $( 9 , - 4 )$ directrix: y=-1
E)vertex: $( 9 , - 4 )$ directrix: y=-7

Question

Quizplus · Accepted Answer

To find the vertex and directrix of the parabola given by $x^2 - 18x - 12y + 33 = 0$, we first complete the square for the x-terms and rearrange the equation to make it resemble the standard form of a parabola equation. 1. Rearrange the equation: $x^2 - 18x = 12y - 33$.2. Complete the square for the x-terms: $x^2 - 18x + 81 = 12y - 33 + 81$, which simplifies to $(x-9)^2 = 12(y+4)$.3. From the completed square form $(x-h)^2 = 4p(y-k)$, where $(h, k)$ is the vertex, we can see that the vertex is $(9, -4)$.4. The value of $4p = 12$, so $p = 3$. Since $p$ is positive, the parabola opens upwards, and the directrix is $k - p = -4 - 3 = -7$.Therefore, the vertex is $(9, -4)$ and the directrix is $y = -7$, which matches choice E.

Find the Vertex and Directrix of the Parabola x2−18x−12y+33=0x ^ { 2 } - 18 x - 12 y + 33 = 0x2−18x−12y+33=0

Find the Vertex and Directrix of the Parabola $x ^ { 2 } - 18 x - 12 y + 33 = 0$