If three normals are drawn from pt. (p,0) to parabola y2=x
then p > 2a where a=1/4.
True or false. and reason ??
19true
also remember that if three normals can be drawn from a pt. (h,k) to the parabola y2=4ax then 27ak2<4(h-2a)3
19let P(h,k) be any pt. and y2=4ax be the given parabola.
the eqn of any normal to the parabola is y=mx-2am-am3
=>am3+m(2a-h) +k =0
this is a cubic and has three roots(say,m1,m2,m3) !if the three normals are real, then the three roots are also real.
thus m12+m22+m32 >0 (for any values of m1,m2,m3)
=>(m1+m2+m3)3 - 2>0
=>(0)2 - 2/a * (2a-h) >0
=>h>2a
66one of the roots of am3 +m(2a-h)=0 is obviously zero. (the k is zero, if the point is on the axis)
so we get
am2 = h-2a
Obviously, h > 2a since m2 is positive for real m.
