Question

Tangent at a point P1 {other than (0, 0)} on the curve y = x3 meets the curve again at P2. The tangent at P2 meets the curve at P3, and so on. Show that the abscissa of P1, P2, P3,. ..., Pn, form a GP. Also find the ratio [area(Δ P1 P2 P3)]/[area(P2 P3 P4)]

Answer 1

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Answer 2

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Let  a point P1 on y = x³ be (h,h³)

Tangent at P1 is y - h³ = 3h²(x-h)...(1)

It meets y = x³ at P2 .

Substituting the value of y in equation (1) :

x³ - h³ = 3h²(x-h)

➠ (x-h)(x²+xh+h²)=3h²(x-h)

➠ x² +xh+h²-3h²=0

➠ x²+xh-2h² = 0

➠(x-h)(x+2h)=0

➠ x = h  or x= -2h

→ x=h is point for P1

→ x = -2h is the point P2.

➠ y = -8h³

Hence, P2 ≡ ( -2h,-8h³)

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)

➠x²-2xh-8h²=0

➠ (x-4h)(x+2h)=0

Hence,

→ x=4h for P3

→x=-2h for P2.

∴ P3 ≡(4h,64h)

Similarly , we get x=-8h for P4.

Hence, the abscissa for P1,P2,P3... are h,-2h,4h...

These are in GP.

❥ Hence proved!

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Now,

$\implies \sf Ar(\triangle P_1 P_2 P_3 ) = 1/2 \: \left | \sf \begin{array}{c c c} h&h^3 &1 \\ -2h&-8h^3&1 \\ 4h&64h^3&1 \end{array} \right |$

And,

$\implies \sf Ar(\triangle P_2 P_3 P_4 )=1/2 \left | \begin{array}{c c c} -2h&-8h^3&1 \\ 4h&64h^3 &1 \\ -8h&-512h^3 &1 \end{array} \right |$

$\implies \sf 1/2 (-2)(-8) \left | \begin{array}{ c c c} h&h^3&1 \\ -2h&-8h^3&1 \\ 4&64h^3&1 \end{array} \right |$

Now,

$\implies \sf \frac{Ar.(\triangle P_1 P_2 P_3 )}{Ar.(\triangle P_2 P_3 P_4)}=\frac{1/2 \left | \begin{array}{ccc} h&h^3&1 \\ -2h&-8h^3&1 \\ 4h&64h^3&1 \end{array}\right |}{1/2 (-2)(-8)\left | \begin{array}{ccc} h&h^3&1 \\ -2h&-8h^3&1 \\ 4h&64h^3&1 \end{array}\right |}$

❥ Ratio of area(Δ P1 P2 P3)]/[area(P2 P3 P4)] = 1/ 16 .

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