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Question

Tangent at a point P1{otherthan(0,0)} on the curve y=x3 meets the curve again at P2. The tangent at P2 meet the curve at P3 and so on. Then the numerical value of [area(ΔP1P2P3)][area(ΔP2P3P4)]

A
1/2
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B
1/4
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C
1/8
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D
1/16
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Solution

The correct option is A 1/2
Let any point P1 on y=x3 be (h,h3)
Then tangent at P1 is yh3=3h2(xh) ...(1)
It meets y=x3 at P2
On putting the value of y in equation (1)
x3h3=3h2(xh)(xh)(x2+xh+h2)=3h2(xh)
x2+xh+h2=3h2 or x=h
x2+xh2h2=0(xh)(x+2h)=0
x=h or x=2h
Therefore, x=2h is the point P2 which implies y=8h3
Hence, point P2(2h,8h2)
Again, tangent at P2 is y+8h3=3(2h)2(x+2h)
It meets y=x3 at P3
x3+8h2=12h2(x+2h)x22hx8h2=0(x+2h)(x4h)=0x=4hy=64h3
Therefore P3(4h,64h3)
Similarly, we get P4(8h,83h3)
Hence, the absissae are h,2h,4h,8h,... which form a GP
Let D=P1P2P3 and D"=P2P3P4
DD"=P1P2P3P2P3P4=12∣ ∣ ∣hh312h8h314h64h31∣ ∣ ∣12∣ ∣ ∣2h8h314h64h318k512h31∣ ∣ ∣
=12∣ ∣ ∣hh312h8h314h64h31∣ ∣ ∣12×(2)×(8)∣ ∣ ∣hh312h8h314h64h31∣ ∣ ∣
C2DD"=116 which is the required ratio

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