Aparabola with a vertex at (0,0) has a directrix that crosses the negative part of the y-axis. which could be the equation of the parabola? x2 = –4y x2 = 4y y2 = 4x y2 = –4x

In the attached files you can see graphs of all parabolas and their directrices.

As you can see the graph of parabola with a vertex at (0,0) and with a directrix that crosses the negative part of the y-axis is represented at the second diagram.

correct choice is B,

Step-by-step explanation:

Since, if the directrix of a parabola crosses the negative part of the y-axis.

Then it is obvious that the parabola must be upward,

Since, the equation of upward parabola is,

Where, a > 1,

Where (h, k) is the vertex of the parabola,

h = 0, k = 0,

Then the equation of the upward parabola,

∵ a > 0 ⇒ 1/a > 0,

Since, -4 < 0,

Hence, the required equation of the parabola would be,

Note : and have the directrix on the negative x-axis and positive x-axis respectively.

$32,660

step-by-step explanation:

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116