The correct options are
A Abscissa of P1,P2,⋯,P3 are in G.P.
B Ordinate of P1,P2,⋯,P3 are in G.P.
D Ratio of area of ΔP1P2P3 and ΔP2P3P4 is 116
Let P1 be (h,h3) on y=x3
dydx=3x2
Tangent at P1 is,
y−h3=3h2(x−h) ⋯(1)
It meets again y=x3 at P2.
Putting the value of y in eqn(1)
x3−h3=3h2(x−h)
⇒(x−h)(x2+xh+h2)−3h2(x−h)=0
⇒(x−h)2(x+2h)=0⇒x=h,−2h
⇒x=−2h is the point P2.
⇒y=−8h3
∴P2≡(−2h,−8h3)
Again, tangent at P2 is,
y+8h3=3(−2h)2(x+2h)
It meets again y=x3 at P3.
⇒x3+8h3=12h2(x+2h)⇒(x+2h)(x2−2hx+4h2)−12h2(x+2h)=0⇒(x+2h)2(x−4h)=0
⇒x=4h,y=64h3
∴P3≡(4h,64h3)
Similarly, P4≡(−23h,−83h3)
Therefore, the abscissa h,−2h,4h,−8h⋯ form a G.P.
The ordinate h3,−8h,82h,−83h⋯ form a G.P.
Let T1=ΔP1P2P3 and T2=ΔP2P3P4
T1T2=ΔP1P2P3ΔP2P3P4=12∣∣
∣
∣∣hh31−2h−8h314h82h31∣∣
∣
∣∣12∣∣
∣
∣∣−2h−8h314h82h31−8h−83h31∣∣
∣
∣∣
=∣∣
∣
∣∣hh31−2h−8h314h82h31∣∣
∣
∣∣(−2)(−8)∣∣
∣
∣∣hh31−2h−8h314h82h31∣∣
∣
∣∣
=116