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Equation of Tangent at a Point (x,y) in Terms of f'(x)

Tangent at a ...

Question

Tangent at a point $P_{1}$ (other then $(0, 0)$ on the curve $y = x^{3}$ meets it again at $P_{2}$ . The tangent at $P_{2}$ meets the curve at $P_{3}$ & so on. Show that the abscissa of $P_{1}, P_{2}, P_{3}, . . . . . . P_{n}$ form a $G . P$ . Also find the ratio $\frac{a r e a o f Δ (P_{1} P_{2} P_{3})}{a r e a o f Δ (P_{2} P_{3} P_{4})}$

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Solution

Let a point $P_{1}$ on $y = x^{3}$ be $(h, h^{3})$ .
Tangent at $P_{1}$ is $y - h^{3} = 3 h^{2} (x - h)$ , it meets $y = x^{3}$ at $P_{2}$ .
$∴ x^{3} - h^{3} = 3 h^{2} (x - h)$
$\Rightarrow (x - h) (x^{2} + x h + h^{2}) = 3 h^{2} (x - h) [∵ a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})]$
$\Rightarrow x^{2} + x h + h^{2} - 3 h^{2}$
$\Rightarrow x^{2} + x h - 2 h^{2} = 0$
$\Rightarrow (x - h) (x + 2 h) = 0$
$∴ x = - 2 h$ for $P_{2}$ $x = h$ is for point $P_{1}$
$∴ P_{2}$ is $(- 2 h, - 8 h^{3})$
Again, tangent at $P_{2}$ is $y + 8 h^{3} = 3 {(2 h)}^{2} (x + 2 h)$ , it meets $y = x^{3}$ at $P_{3}$ .
$∴ x^{3} + 8 h^{3} = 12 h^{2} (x + 2 h)$
$\Rightarrow (x + 2 h) (x^{2} - 2 x h + 4 h^{2}) = 12 h^{2} (x - h) [∵ a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})]$
$\Rightarrow x^{2} - 2 x h + 4 h^{2} - 12 h^{2}$
$\Rightarrow x^{2} - 2 x h - 8 h^{2} = 0$
$\Rightarrow (x - 4 h) (x + 2 h) = 0$
$∴ x = 4 h$ for $P_{3}$ $x = - 2 h$ is for point $P_{2}$
$∴ P_{3}$ is $(4 h, 64 h^{3})$
Similarly, we get
$x = - 8 h$ for $P_{4} . . . . . . . .$
Hence, the abscissa for $P_{1}, P_{2}, P_{3}, . . . . . . .$ are $h, - 2 h, 4 h, . . . . . . .$ respectively, which are in G.P.. [Hence proved]
Now for $△ (P_{1} P_{2} P_{3})$
$a r (△ (P_{1} P_{2} P_{3})) = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} h & h^{3} & 1 - 2 h & - 8 h^{3} & 1 4 h & 64 h^{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
Again for $△ (P_{2} P_{3} P_{4})$
$a r (△ (P_{2} P_{3} P_{4})) = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} - 2 h & - 8 h^{3} & 1 4 h & 64 h^{3} & 1 - 8 h & - 512 h^{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = \frac{1}{2} (- 2) (- 8) ∣ ∣ ∣ ∣ ∣ \begin{matrix} h & h^{3} & 1 - 2 h & - 8 h^{3} & 1 4 h & 64 h^{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$∴ \frac{a r (△ (P_{1} P_{2} P_{3}))}{a r (△ (P_{2} P_{3} P_{4}))} = \frac{\frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} h & h^{3} & 1 - 2 h & - 8 h^{3} & 1 4 h & 64 h^{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣}{\frac{1}{2} (- 2) (- 8) ∣ ∣ ∣ ∣ ∣ \begin{matrix} h & h^{3} & 1 - 2 h & - 8 h^{3} & 1 4 h & 64 h^{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣}$
$\Rightarrow \frac{a r (△ (P_{1} P_{2} P_{3}))}{a r (△ (P_{2} P_{3} P_{4}))} = \frac{1}{16}$

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