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# Tangent at a point P1 (other then (0,0) on the curve y=x3 meets it again at P2. The tangent at P2 meets the curve at P3 & so on. Show that the abscissa of P1,P2,P3,......Pn form a G.P. Also find the ratio areaofΔ(P1P2P3)areaofΔ(P2P3P4)

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## Let a point P1 on y=x3 be (h,h3).Tangent at P1 is y−h3=3h2(x−h), it meets y=x3 at P2.∴x3−h3=3h2(x−h)⇒(x−h)(x2+xh+h2)=3h2(x−h)[∵a3−b3=(a−b)(a2+ab+b2)]⇒x2+xh+h2−3h2⇒x2+xh−2h2=0⇒(x−h)(x+2h)=0∴x=−2h for P2 x=h is for point P1∴P2 is (−2h,−8h3)Again, tangent at P2 is y+8h3=3(2h)2(x+2h), it meets y=x3 at P3.∴x3+8h3=12h2(x+2h)⇒(x+2h)(x2−2xh+4h2)=12h2(x−h)[∵a3+b3=(a+b)(a2−ab+b2)]⇒x2−2xh+4h2−12h2⇒x2−2xh−8h2=0⇒(x−4h)(x+2h)=0∴x=4h for P3 x=−2h is for point P2∴P3 is (4h,64h3)Similarly, we getx=−8h for P4........Hence, the abscissa for P1,P2,P3,....... are h,−2h,4h,....... respectively, which are in G.P.. [Hence proved]Now for △(P1P2P3)ar(△(P1P2P3))=12∣∣ ∣ ∣∣hh31−2h−8h314h64h31∣∣ ∣ ∣∣Again for △(P2P3P4)ar(△(P2P3P4))=12∣∣ ∣ ∣∣−2h−8h314h64h31−8h−512h31∣∣ ∣ ∣∣=12(−2)(−8)∣∣ ∣ ∣∣hh31−2h−8h314h64h31∣∣ ∣ ∣∣∴ar(△(P1P2P3))ar(△(P2P3P4))=12∣∣ ∣ ∣∣hh31−2h−8h314h64h31∣∣ ∣ ∣∣12(−2)(−8)∣∣ ∣ ∣∣hh31−2h−8h314h64h31∣∣ ∣ ∣∣⇒ar(△(P1P2P3))ar(△(P2P3P4))=116  Suggest Corrections  0      Similar questions
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