document.write( "Question 719601: Find the coordinates of the vertex,focus,ends of the latus rectum and the equation of the directrix.Draw the parabola of:\r
\n" ); document.write( "\n" ); document.write( "x^2=12(y+7) \n" ); document.write( "

Algebra.Com's Answer #442045 by lwsshak3(11628) $\"\"$ $\"About$

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Find the coordinates of the vertex,focus,ends of the latus rectum and the equation of the directrix.Draw the parabola of:
\n" ); document.write( "x^2=12(y+7)
\n" ); document.write( "...
\n" ); document.write( "This is an equation of a parabola that opens upwards:
\n" ); document.write( "Its basic form of equation:(x-h)^2=4p(y-k)
\n" ); document.write( "For given equation: x^2=12(y+7)
\n" ); document.write( "vertex: (0,-7)
\n" ); document.write( "axis of symmetry: x=0
\n" ); document.write( "4p=12
\n" ); document.write( "p=3
\n" ); document.write( "focus: (0,-4) (p-units above vertex on the axis of symmetry)
\n" ); document.write( "Ends of latus rectum(focal width):
\n" ); document.write( "plugs in y-coordinate of focus(-4) then solve for x
\n" ); document.write( "x^2=12(y+7)
\n" ); document.write( "x^2=12(-4+7)=12*3=36
\n" ); document.write( "x=ą√36=ą6
\n" ); document.write( "ends of latus rectum: (-6,-4) and (6,-4)\r
\n" ); document.write( "\n" ); document.write( "see graph below:\r
\n" ); document.write( "\n" ); document.write( " $\"+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+%28x%5E2-84%29%2F12%29\"$ \n" ); document.write( "

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