Question

Which of the following equations is of the parabola whose vertex is at (2, 3), axis of symmetry parallel to the y-axis and p = 4?
A.)y-3 = 1/16 (x-2)^2
B.)y+3 = -1/16 (x+2)^2
C.)x-2 = 1/16 (y-3)^2

Answer 1

Answer:

option A.) y-3 = 1/16 (x-2)^2

Step-by-step explanation:

we know that

If the axis of symmetry is parallel to the y-axis, then we have a vertical parabola

The equation of a vertical parabola is equal to

$4p(y-k)=(x-h)^{2}$

where

(h,k) is the vertex

In this problem we have

(h,k)=(2,3)

p=4

substitute

$4(4)(y-3)=4(4)(x-2)^{2}$

$16(y-3)=(x-2)^{2}$

$(y-3)=(1/16)(x-2)^{2}$