N nrlwinner Member Joined Apr 18, 2009 Messages 194 Gender Male HSC 2010 Feb 11, 2010 #1 How can I solve that to get to
shaon0 ... Joined Mar 26, 2008 Messages 2,029 Location Guess Gender Male HSC 2009 Feb 11, 2010 #2 nrlwinner said: How can I solve that to get to Click to expand... x=(t-(4/t)) x^2=(t-(4/t))^2 x^2=(t^2+(16/t^2))-8 x^2+8=t^2+16/t^2 (x^2+8)/2=(t^2)/2+8/t^2 (x^2+8)/2=y x^2+8=2y x^2=2(y-4)
nrlwinner said: How can I solve that to get to Click to expand... x=(t-(4/t)) x^2=(t-(4/t))^2 x^2=(t^2+(16/t^2))-8 x^2+8=t^2+16/t^2 (x^2+8)/2=(t^2)/2+8/t^2 (x^2+8)/2=y x^2+8=2y x^2=2(y-4)
N nrlwinner Member Joined Apr 18, 2009 Messages 194 Gender Male HSC 2010 Feb 12, 2010 #3 Got another Q here. and are two points on the parabola . The chord PQ subtends a right angle at the origin. If R is the fourth vertex of the rectangle POQR show that R parametrically is when st=-4
Got another Q here. and are two points on the parabola . The chord PQ subtends a right angle at the origin. If R is the fourth vertex of the rectangle POQR show that R parametrically is when st=-4
E elv09 Member Joined May 2, 2009 Messages 44 Gender Male HSC 2009 Feb 18, 2010 #5 http://www.easy-share.com/1909317349/IMG_0006.jpg i could be wrong for the x-coordinate ...