An equation of a parabola is given. Write the equation of the parabola in standard form, and identify the
vertex and the focus.
-$$x ^ { 2 } - 8 x - 16 y - 16 = 0$$
A) $( x - 4 ) ^ { 2 } = 16 ( y + 2 )$
vertex: $( - 2,4 )$; focus $( 2,4 )$
B) $( x - 4 ) ^ { 2 } = 16 ( y + 2 )$
vertex: $( - 4,2 )$; focus $( - 4,6 )$
C) $( x + 4 ) ^ { 2 } = 16 ( y - 2 )$
vertex: $( - 4,2 )$; focus $( - 4,6 )$
D) $( x - 4 ) ^ { 2 } = 16 ( y + 2 )$
vertex: $( 4 , - 2 )$; focus $( 4,2 )$

Question

Quizplus · Accepted Answer

The equation can be rewritten in standard form by completing the square on the x-terms. The correct standard form is $(x - 4)^2 = 16(y + 2)$, indicating a parabola that opens upwards with vertex at $(4, -2)$. The focus is located at a distance $\frac{1}{4} \times 16 = 4$ units above the vertex, making it $(4, 2)$.

An Equation of a Parabola Is Given x2−8x−16y−16=0x ^ { 2 } - 8 x - 16 y - 16 = 0x2−8x−16y−16=0

An Equation of a Parabola Is Given $x ^ { 2 } - 8 x - 16 y - 16 = 0$