If normals drawn at three different point on the parabola `y^(2)=4ax` pass through the point (h,k), then show that h `hgt2a`.

Explanation

The equation of a normal to the parabola `y^(2)=4ax` having slope m is
`y=mx-2am-am^(3)` (1)
Since the normal drawn at three different points on the parabola pass through (h,k), it must satisfy the equation (1).
`:." "k=mh-2am-am^(3)`
`rArr" "am^(3)+(2a-h)m+k=0` (2)
Now, it is given that three different normals pass through (h,k).
Thus, equation (2) has three distinct real roots `m_(1),m_(2)andm_(3)`.
From equation (2),
`m_(1)+m_(2)+m_(3)=0` (3) and `m_(1)m_(2)+m_(2)m_(3)+m_(3)m_(1)=2a-h` (4)
`rArr" "m_(1)^(2)+m_(2)^(2)+m_(3)^(2)=-(m_(1)m_(2)+m_(2)m_(3)+m_(3)m_(1))`
`rArr" "m_(1)^(2)+m_(2)^(2)+m_(3)^(2)=2(h-2a)`
Since `m_(1),m_(2)andm_(3)` are distinct and real, L.H.S `gt0`
(As in case of imaginary roots L.H.S. may be negative)
`:." "h-2agt0`
`rArr" "hgt2a`
Alternative method :
Let `am^(2)+(2a-h)m+k=f(m)`
Since equation (2), i.e., f(m)=0 has three distinct real roots, then equation f' (m) = 0 must have distinct real roots.

`:." "3am^(2)+(2a-h)=0` has two distinct real roots.
So, `2a-hgt0orhgt2a`.
Note: The converse of this is not true, i.e., if `h lt 2a` then it is not necessary that equation (2) has three distinct real roots. In other words, if `h lt 2a` then it is not necessary that three distinct normals can be drawn from (h, k).