wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Tangent at a point P1 (other than origin) on the curve y=x3 meets the curve again at P2. The tangent at P2 meets the curve at P3 and so on. Then which of the following is/are correct?

A
Abscissa of P1,P2,,P3 are in G.P.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Ordinate of P1,P2,,P3 are in G.P.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Ratio of area of ΔP1P2P3 and ΔP2P3P4 is 18
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Ratio of area of ΔP1P2P3 and ΔP2P3P4 is 116
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

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,
yh3=3h2(xh) (1)
It meets again y=x3 at P2.
Putting the value of y in eqn(1)
x3h3=3h2(xh)
(xh)(x2+xh+h2)3h2(xh)=0
(xh)2(x+2h)=0x=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)(x22hx+4h2)12h2(x+2h)=0(x+2h)2(x4h)=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∣ ∣ ∣hh312h8h314h82h31∣ ∣ ∣12∣ ∣ ∣2h8h314h82h318h83h31∣ ∣ ∣
=∣ ∣ ∣hh312h8h314h82h31∣ ∣ ∣(2)(8)∣ ∣ ∣hh312h8h314h82h31∣ ∣ ∣
=116

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Applications
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon