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cutiecomittee123

  • one year ago

How do I find the directrix of this equation y^2+4x+4y-4=0

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  1. anonymous
    • one year ago
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    First you would need to complete the square on the y-terms to help you put the parabola into a standard form. Are you okay on completing the square?

  2. cutiecomittee123
    • one year ago
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    (y^2+4y+ 4 )+ (4x) =4

  3. anonymous
    • one year ago
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    RIght, looks like you get the idea. Okay, we would have \((y+2)^{2} = 4-4x\) \((y+2)^{2} = -4(x-1)\) In order to get it into standard form, I factored out a negative 4 instead of simply a positive 4. So now we have a form we can use that matches up with \((y-k)^{2} = -4p(x-h)\) So when we have this specific form, where there is a negative in front and the y is the variable being squared, we have a parabola opening to the left. This means that the directrix will be to the right of the vertex of our parabola. As you probably know, (h,K) is the vertex of the parabola, so for us we have our vertex at (1,-2) With me so far?

  4. cutiecomittee123
    • one year ago
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    got it:)

  5. anonymous
    • one year ago
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    Okay, cool. So now the value in front of (x-h) portion is equal to 4p (we dont need to worry about the sign, we already know we need the directrix to be to the right of the vertex). So that means 4p = 4 and thus p = 1. This value of p helps us get everything else that we need to graph. From the vertex, the focus is a distance of p away while inside the bowl of the parabola and the directrix is a distance of p from the vertex in the opposite direction. So we only need to go one to the right of our vertex. Meaning the directrix is the line x = 2. Just make sure that you note that the directrix is a line and not just a number.

  6. cutiecomittee123
    • one year ago
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    okay that makes sense

  7. cutiecomittee123
    • one year ago
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    Thanks:)

  8. anonymous
    • one year ago
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    No problem :)

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