Normals at `(x_(1),y_(1)),(x_(2),y_(2))and(x_(3),y_(3))` to the parabola `y^(2)=4x` are concurrent at point P. If `y_(1)y_(2)+y_(2)y_(3)+y_(3)y_(1)=x_

Correct Answer - 4
4 Equation of normal to parabola `y^(2)=4x` having slope m is
`y=mx-2m-m^(3)`
This normal passes through the point P(h,k).
`:." "k=mh-2m-m^(3)`
`or" "m^(3)+(2-h)m+k=0` (1)
This equation has three real roots `m_(1),m_(2)andm_(3)`, which are slope of three normals.
Also, three feet of normals on the plane are `(m_(1)^(2),-2m_(1)),(m_(2)^(2),-2m_(2))and(m_(3)^(2),-2m_(3))`.
Given that
`y_(1)y_(2)+y_(2)y_(3)+y_(3)y_(1)=x_(1)x_(2)x_(3)`
`:." "4m_(1)m_(2)+4m_(2)m_(3)+4m_(3)m_(1)=m_(1)^(2)m_(2)^(2)m_(3)^(2)` (2)
From equation (1), we have
`m_(1)m_(2)+m_(2)m_(3)+m_(3)m_(1)=2-h`
`andm_(1)m_(2)m_(3)=-k`
Putting these value in (2), we get
`4(2-h)=(-k)^(2)`
`:." "y^(2)=-4(x-2)`, which is locus of point P.
This equation shows a parabola with latus rectum length 4 units.