log (1 Viewer)

[fullonoob]

equation: x^2 + 4cx/(1+c^2) +1 = 0
gradient of log(1+x^2) a x=c is y' = 2c / (1+c^2)

Show that delta = -4 (1-c^2 / 1+c^2)^2 for this quadratic, and hence that the only tangents to y = log (1+x^2) which are mutually perpendicular are those at x= -1 and x= 1.

skip the show part and help me with the show x= -1 and x = 1 please

 

[shaon0]

equation: x^2 + 4cx/(1+c^2) +1 = 0
gradient of log(1+x^2) a x=c is y' = 2c / (1+c^2)

Show that delta = -4 (1-c^2 / 1+c^2)^2 for this quadratic, and hence that the only tangents to y = log (1+x^2) which are mutually perpendicular are those at x= -1 and x= 1.

skip the show part and help me with the show x= -1 and x = 1 please

If c=+-1. Delta=0 [you want Delta=0 so there's a kind of double root for log(1+x^2) so tangents only touch once]
ie x=-4c/(1+c^2)/2
=-2/2 or +2/2
=+-1
And, y'=+-1 ie at x=+-1 the tangents are mutually perpendicular.

 

[katietheskatie]

[youtube]zUnGXTjmc9E[/youtube]

just sayin'

 

[fullonoob]

lols just forgot the delta = 0 part xD ty

 

[hello-there]

equation: x^2 + 4cx/(1+c^2) +1 = 0
gradient of log(1+x^2) a x=c is y' = 2c / (1+c^2)

Show that delta = -4 (1-c^2 / 1+c^2)^2 for this quadratic, and hence that the only tangents to y = log (1+x^2) which are mutually perpendicular are those at x= -1 and x= 1.

skip the show part and help me with the show x= -1 and x = 1 please

lol do the show part im struggling here