DR STRANGE
Last Activity: 8 Years ago
equation of normal to parabola y2=4x is y=mx-2m-m3 it passes through ( h,k)
hence ,k =mh -2m-m3
m3+(2-h)m +k=0
let m1 , m2 ,m3 be roots of the equation , now we will use properties cubic equations,
then m1+m2+m3= -coefficient of m2/ coefficient of m3 =0 …........(as there is no m2 term)
m1m2+m2m3+m3m1=coefficient of m/ coefficient of m3 = 2-h
m1m2m3 = -constant term/ coefficient of m3 = -k ,hence all the values m1 ,m2 , m3 cant be zero
m12+m22+m32 >0
i.e. (m1+m2+m3)2 -2(m1m2+m2m3+m3m1)>0
i.e. ( 0 )2 -2 (2-h)>0
i.e -2(2-h)>0
i.e. h-2>0 => h>2 …..........................result will be same for y2=4ax
please approve if you are satisfied :)