SATHEE: Area Question 33

Area Question 33

Question 33

A curve passes through $(2, 0)$ and the slope of tangent at point $P (x, y)$ equals $\frac{(x + 1)^{2} + y - 3}{(x + 1)}$ .

Find the equation of the curve and area enclosed by the curve and the $X$ -axis in the fourth quadrant. (2004, 5M)

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Solution:

Here, slope of tangent,

$\begin{aligned} \frac{d y}{d x} & = \frac{(x + 1)^{2} + y - 3}{(x + 1)} \Rightarrow \frac{d y}{d x} & = (x + 1) + \frac{(y - 3)}{(x + 1)}, \end{aligned}$

Put $x + 1 = X$ and $y - 3 = Y$

$\begin{array}{llrl} \frac{d y}{d x} & = \frac{d Y}{d X} & \frac{d Y}{d X} & = X + \frac{Y}{X} & \frac{d Y}{d X} & = \frac{1}{X} Y = X & IF = e^{\int - \frac{1}{X} d X} & = e^{- \log X} = \frac{1}{X} \end{array}$

$∴$ Solution is, $Y \cdot \frac{1}{X} = \int X \cdot \frac{1}{X} d X + c$

$\Rightarrow \frac{Y}{X} = X + c$

$y - 3 = (x + 1)^{2} + c (x + 1)$ , which passes through $(2, 0)$ .

$\Rightarrow - 3 = (3)^{2} + 3 c$

$\Rightarrow c = - 4$

$∴$ Required curve

$\begin{aligned} y & = (x + 1)^{2} - 4 (x + 1) + 3 \Rightarrow & y & = x^{2} - 2 x \end{aligned}$

$∴$ Required area $= \int_{0}^{2} (x^{2} - 2 x) d x = \frac{x^{3}}{3} - x^{2}_{0}^{2}$

$= \frac{8}{3} - 4 = \frac{4}{3} sq units$

Area Question 33

Question 33

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Area Question 33

Question 33

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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