SATHEE: Application of Derivatives 1 Question 19

Application of Derivatives 1 Question 19

####19. Tangent at a point $P_{1}$ {other than $Missing or unrecognized delimiter for \right$ on the curve $y = x^{3}$ meets the curve again at $P_{2}$ . The tangent at $P_{2}$ meets the curve at $P_{3}$ and so on.

Show that the abscissa of $P_{1}, P_{2}, P_{3}, \dots, P_{n}$ , form a GP. Also, find the ratio of

[area $(Δ P_{1} P_{2} P_{3})] / [area (Δ P_{2} P_{3} P_{4})]$ .

(1993, 5M)

Show Answer

Answer:

( $1 : 16$ )

Solution:

Let any point $P_{1}$ on $y = x^{3}$ be $(h, h^{3})$ .

Then, tangent at $P_{1}$ is

It meets $y = x^{3}$ at $P_{2}$ .

$y - h^{3} = 3 h^{2} (x - h)$

On putting the value of $y$ in Eq. (i), we get

$\begin{aligned} x^{3} - h^{3} = 3 h^{2} (x - h) \\ \Rightarrow (x - h) (x^{2} + x h + h^{2}) = 3 h^{2} (x - h) \\ \Rightarrow x^{2} + x h + h^{2} = 3 h^{2} or x = h \\ \Rightarrow x^{2} + x h - 2 h^{2} = 0 \\ \Rightarrow (x - h) (x + 2 h) = 0 \\ \Rightarrow x = h or x = - 2 h \end{aligned}$

Therefore, $x = - 2 h$ is the point $P_{2}$ ,

which implies $y = - 8 h^{3}$

Hence, point $P_{2} \equiv (- 2 h, - 8 h^{3})$

Again, tangent at $P_{2}$ is $y + 8 h^{3} = 3 (- 2 h)^{2} (x + 2 h)$ .

$\begin{array}{lrl} It meets & y & = x^{3} at P_{3} \\ \Rightarrow x^{3} + 8 h^{3} & = 12 h^{2} (x + 2 h) \\ \Rightarrow x^{2} - 2 h x - 8 h^{2} & = 0 \\ \Rightarrow (x + 2 h) (x - 4 h) & = 0 \Rightarrow x = 4 h \Rightarrow y = 64 h^{3} \\ Therefore, P_{3} & \equiv (4 h, 64 h^{3}) \\ Similarly, we get P_{4} & \equiv (- 8 h, - 8^{3} h^{3}) \end{array}$

Hence, the abscissae are $h, - 2 h, 4 h, - 8 h, \dots$ , which form a GP.

Let $D^{'} = Δ P_{1} P_{2} P_{3}$ and $D^{''} = Δ P_{2} P_{3} P_{4}$

$\begin{aligned} \frac{D^{'}}{D^{''}} & = \frac{Δ P_{1} P_{2} P_{3}}{Δ P_{2} P_{3} P_{4}} = \frac{\frac{1}{2} | \begin{array}{ccc} h & h^{3} & 1 \\ - 2 h & - 8 h^{3} & 1 \\ 4 h & 64 h^{3} & 1 \end{array} |}{\frac{1}{2} | \begin{array}{ccc} - 2 h & - 8 h^{3} & 1 \\ 4 h & 64 h^{3} & 1 \\ - 8 h & - 512 h^{3} & 1 \end{array} |} \\ = \frac{\frac{1}{2} | \begin{array}{ccc} h & h^{3} & 1 \\ - 2 h & - 8 h^{3} & 1 \\ 4 h & 64 h^{3} & 1 \end{array} |}{\frac{1}{2} \times (- 2) \times (- 8) | \begin{array}{ccc} h & h^{3} & 1 \\ - 2 h & - 8 h^{3} & 1 \\ 4 h & 64 h^{3} & 1 \end{array} |} \end{aligned}$

$\Rightarrow \frac{D^{'}}{D^{'}} = \frac{1}{16} = 1 : 16$ which is the required ratio.

Application of Derivatives 1 Question 19

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Application of Derivatives 1 Question 19

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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