# directrix of y=-4

• Apr 19th 2009, 04:55 PM
peekybooyou
directrix of y=-4
have a vertex at (-4,-1) with a vertical axis of symennetry. which one has a directrix of y=-4

so far i got (x+4)^2=?(y+1)

what is the ?..?!! how do i get that
• Apr 19th 2009, 10:51 PM
Directrix of a parabola
Hello peekybooyou
Quote:

Originally Posted by peekybooyou
have a vertex at (-4,-1) with a vertical axis of symennetry. which one has a directrix of y=-4

so far i got (x+4)^2=?(y+1)

what is the ?..?!! how do i get that

You're right so far. You have used the fact that the parabola \$\displaystyle X^2 = 4aY\$ has vertex at \$\displaystyle (0,0)\$ and a vertical axis of symmetry. You've replaced \$\displaystyle X\$ by \$\displaystyle (x+4)\$ and \$\displaystyle Y\$ by \$\displaystyle (y+1)\$ to give a parabola whose vertex is at \$\displaystyle (-4,-1)\$.

Now you need to use the fact that the parabola \$\displaystyle X^2 = 4aY\$ has directrix \$\displaystyle Y = -a\$.

If in the \$\displaystyle (x,y)\$ coordinates, the directrix is \$\displaystyle y = -4\$, in the \$\displaystyle (X,Y)\$ coordinates it is

\$\displaystyle Y = y+1 =-3\$

\$\displaystyle \Rightarrow a = 3\$

So in \$\displaystyle (X,Y)\$ coordinates the equation is \$\displaystyle X^2 = 12 Y\$; in \$\displaystyle (x,y)\$ coordinates it is \$\displaystyle (x+4)^2 = 12(y+1)\$