Find the vertex,focus,and directrix of the parabola. $( x + 3 ) + ( y - 1 ) ^ { 2 } = 0$
A)Vertex: $( - 3,1 )$ ;Focus: $\left( \frac { - 13 } { 4 } , 1 \right)$ ;Directrix: $x = \frac { - 11 } { 4 }$
B)Vertex: $( 1 , - 3 , )$ ;Focus: $\left( 1 , \frac { - 11 } { 4 } \right)$ ;Directrix: $x = \frac { - 13 } { 4 }$
C)Vertex: $( 1 , - 3 , )$ ;Focus: $\left( \frac { - 11 } { 4 } , 1 \right)$ ;Directrix: $x = \frac { - 13 } { 4 }$
D)Vertex: $( - 3,1 )$ ;Focus: $\left( 1 , \frac { - 13 } { 4 } \right)$ ;Directrix: $x = \frac { - 13 } { 4 }$
E)Vertex: $( - 3,1 )$ ;Focus: $\left( 1 , \frac { - 11 } { 4 } \right)$ ;Directrix: $x = \frac { - 13 } { 4 }$

Question

Find the vertex,focus,and directrix of the parabola.​ $( x + 3 ) + ( y - 1 ) ^ { 2 } = 0$ ​
A)Vertex: $( - 3,1 )$ ;Focus: $\left( \frac { - 13 } { 4 } , 1 \right)$ ;Directrix: $x = \frac { - 11 } { 4 }$
B)Vertex: $( 1 , - 3 , )$ ;Focus: $\left( 1 , \frac { - 11 } { 4 } \right)$ ;Directrix: $x = \frac { - 13 } { 4 }$
C)Vertex: $( 1 , - 3 , )$ ;Focus: $\left( \frac { - 11 } { 4 } , 1 \right)$ ;Directrix: $x = \frac { - 13 } { 4 }$
D)Vertex: $( - 3,1 )$ ;Focus: $\left( 1 , \frac { - 13 } { 4 } \right)$ ;Directrix: $x = \frac { - 13 } { 4 }$
E)Vertex: $( - 3,1 )$ ;Focus: $\left( 1 , \frac { - 11 } { 4 } \right)$ ;Directrix: $x = \frac { - 13 } { 4 }$

Quizplus · Accepted Answer

The correct equation of the parabola is $(x + 3)^2 + (y - 1)^2 = 0$, which seems to be incorrectly transcribed. However, assuming the intended equation is $y = (x + 3)^2 + 1$, the vertex is at $(-3, 1)$. For a parabola in the form $y = a(x - h)^2 + k$, the vertex is $(h, k)$, making $(-3, 1)$ the correct vertex. The focus and directrix cannot be accurately determined from the provided equation due to a transcription error.

Find the Vertex,focus,and Directrix of the Parabola (x+3)+(y−1)2=0( x + 3 ) + ( y - 1 ) ^ { 2 } = 0(x+3)+(y−1)2=0

Find the Vertex,focus,and Directrix of the Parabola $( x + 3 ) + ( y - 1 ) ^ { 2 } = 0$