Find the equation for the parabola that has its focus on the positive xaxis, 4 units away from the directrix.
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Find the equation for the parabola that has its focus on the positive xaxis, 4 units away from the directrix.
Hello, uselessjack!
Is there more information?
There are four forms of the parabola, each with its own equation.
I'll derive one of them . . .
This is a "vertical" parabola, opening upward.Quote:
Find the equation for the parabola that has its focus on the positive xaxis,
4 units away from the directrix.
The focus is at $\displaystyle F(f,0)$
The directrix is: $\displaystyle y = 4$Code:

*  *
 F
  * +    o     *  
*  (f,0) *
* : *
 * : *
 o V
 (f,2)
 :
 :
   +    +     
4

The equation has the form: .$\displaystyle (xh)^2 \:=\:4p(yk)$
The vertex is at: .$\displaystyle V(f,2)$
The value of $\displaystyle p$ is: .$\displaystyle p \,=\,2$
The equation is: .$\displaystyle (xf)^2 \:=\:4(2)\left(y[2]\right) \quad\Rightarrow\quad (xf)^2 \:=\:8(y+2)$
Thank you for your help and time. I am left with confusion still. I thought there would be a specific answer. I am given five choices and still can't figure out which one is correct.
4y^2 = 4x
y^2 = 4x
y^2 = 8x
x^2 = 16y
4y^2 = x