document.write( "Question 968532: Write the equation of the directix of the conic section shown below:
\n" ); document.write( "y^2-4x+4y-4=0\r
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Algebra.Com's Answer #591843 by josgarithmetic(39621) $\"\"$ $\"About$

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Skipping the fundamental derivation, but for directrix (-p,y) using p as a positive number, focus (p,0), for parabola in standard form a axis of symmetry being the x-axis, equation is $\"y%5E2=4px\"$ . If vertex were some point (h,k), then the equation of this parabola is $\"%28y-k%29%5E2=4p%28x-h%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Your example problem equation is easily transformed into $\"%28y%2B2%29%5E2=4%28x%2B2%29\"$ . You can make the correspondances to find p=1 IF you had parabola in standard form and standard position, but your case is that vertex is pushed to the left by 2 units. See, your vertex is (-2,-2); your directrix is $\"x=-1-2\"$ , or simply now $\"highlight%28x=-3%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Remember, p was taken as nonnegative in the fundamental derivation for standard position, so directrix is p units to the left of the vertex point, and vertex would be on the origin. \n" ); document.write( "

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