    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.
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B
Ordinate of P1,P2,,P3 are in G.P.
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C
Ratio of area of ΔP1P2P3 and ΔP2P3P4 is 18
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D
Ratio of area of ΔP1P2P3 and ΔP2P3P4 is 116
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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, 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  Suggest Corrections  0      Similar questions  Related Videos   Applications
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