document.write( "Question 571627: how do i classify conic sectins n write its equation in standard form?
\n" ); document.write( "example:
\n" ); document.write( "-2y^2+x-20y-49=0 \n" ); document.write( "

Algebra.Com's Answer #368148 by KMST(5328) $\"\"$ $\"About$

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The conic sections you are likely to encounter will have axes of symmetry parallel to the x and y axes, and that will make it easier, because you will not see a term with xy.
\n" ); document.write( "Getting to the standard form involves completing squares. Whenever you see a variable and a variable squared, as in
\n" ); document.write( " $\"-2y%5E2-20y=-2%28y%5E2%2B10y%29\"$ you have to imagine the part of the expression with that variable as part of a perfect square.
\n" ); document.write( "In this case
\n" ); document.write( " $\"%28y%2B5%29%5E2=y%5E2%2B10y%2B25\"$ should come to mind.
\n" ); document.write( "With that in mind, you start transforming the equation
\n" ); document.write( " $\"-2y%5E2%2Bx-20y-49=0\"$ --> $\"-2y%5E2-20y%2Bx-49=0\"$ --> $\"-2%28y%5E2%2B10y%29%2Bx-49=0\"$ --> $\"-2%28y%5E2%2B10y%2B25-25%29%2Bx-49=0\"$ --> $\"-2%28%28y%2B5%29%5E2-25%29%2Bx-49=0\"$ --> $\"-2%28y%2B5%29%5E2%2B50%2Bx-49=0\"$ --> $\"-2%28y%2B5%29%5E2%2Bx%2B1=0\"$ --> $\"2%28y%2B5%29%5E2=x%2B1\"$ --> $\"%28y%2B5%29%5E2=%281%2F2%29%28x%2B1%29\"$
\n" ); document.write( "The last equation tells you that it is a parabola with horizontal axis of symmetry $\"y=-5\"$ , vertex at (-1,-5), and focal distance $\"%281%2F2%29%281%2F4%29=1%2F8\"$
\n" ); document.write( "It opens to the right. $\"x%2B1%3E=0\"$ ,--> $\"x%3E=-1\"$ .
\n" ); document.write( "The directrix is the vertical line $\"x=-1-1%2F8=-9%2F8=-1.125\"$ .
\n" ); document.write( "The focus has $\"x=-1%2B1%2F8=-7%2F8=-0.875\"$ , so it's the point(-0.875,-5).
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