document.write( "Question 877244: Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = -1 \n" ); document.write( "
Algebra.Com's Answer #529250 by nerdybill(7384)![]() ![]() You can put this solution on YOUR website! Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = -1 \n" ); document.write( ". \n" ); document.write( "vertex is mid-way between the focus and directrix: \n" ); document.write( "vertex: (-5,2) \n" ); document.write( "standard form of a vertical parabola \n" ); document.write( "(x – h)^2 = 4p(y – k) \n" ); document.write( "(x – (-5))^2 = 4p(y – 2) \n" ); document.write( "(x + 5)^2 = 4p(y – 2) \n" ); document.write( ". \n" ); document.write( "p is the distance between the vertex and either the focus or directrix: \n" ); document.write( "p = 3 \n" ); document.write( ". \n" ); document.write( "so our final equation is: \n" ); document.write( "(x + 5)^2 = 4*3(y – 2) \n" ); document.write( "(x + 5)^2 = 12(y – 2)\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |