If the normals from any point on the parabola `x^2=4y` cut the line y = 2 in points whose abscissa are in A.P., then the slopes of tangents at the 3 co-normal points are :

Options

A. AP
B. GP
C. HP
D. none of these

Answer

B

Explanation

The equation of the at any point `(2t, t^(2))` to the parabola `x^(3)=4y`, is
`x+by=2t+t^(3)" …(i)"`
If it passes through `P(x_(1), y_(2))`, then
`x_(1)+ty_(1)=2t+t_(3)rArrt^(3)+t(2-y_(1))+x_(1)=0`
This is a cubic equation in t. So, it gives three values of t, say `t_(1), t_(2), t_(3)`.
`:." "t_(1)+t_(2)+t_(3)=-("Coff. of "t^(2))/("Coff. of "t^(3))=0`
`rArr" "t_(1)^(3)+t_(2)^(3)+t_(3)^(2)=3t_(1)t_(2)t_(3)" ....(ii)"`
The coordinates of the foot the normals are
`(2t_(1),t_(1)^(2)),(2t_(2), t_(2)^(2))" and "(2t_(3), t_(3)^(2)).`
Thus, three normals from `P(x_(1), y_(1))` cut line y = 2 at `t_(1)^(3), t_(2)^(3)" and "t_(2)^(3)`.
It is given that `t_(1)^(3),t_(2)^(3),t_(3)^(3)" are in AP".`
`:." "2t_(2)^(3)=t_(1)^(3)+t_(3)^(3)`
`rArr" "3t_(2)^(3)=t_(1)^(3)+t_(2)^(3)+t_(2)^(3)`
`rArr" "3t_(2)^(3)=3t_(1)t_(2)t_(3)" "["Using (ii)"]`
`rArr" "t_(2)^(2)=t_(2)t_(3)rArr1/t_(1)xx1/t_(3)=(1/t_(2))^(2)rArr1/t_(1),1/t_(2),1/t_(3)" are in G.P."`
`rArr" Slopes of the tangents at 3 conormal points are in G.P."`