SATHEE: Area Question 10

Area Question 10

Question 10

The area (in sq units) in the first quadrant bounded by the parabola, $y = x^{2} + 1$ , the tangent to it at the point $(2, 5)$ and the coordinate axes is (2019 Main, 11 Jan II) (a) $\frac{14}{3}$ (b) $\frac{187}{24}$ (c) $\frac{8}{3}$ (d) $\frac{37}{24}$

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Solution:

Given, equation of parabola is $y = x^{2} + 1$ , which can be written as $x^{2} = (y - 1)$ . Clearly, vertex of parabola is $(0, 1)$ and it will open upward.

Now, equation of tangent at $(2, 5)$ is $\frac{y + 5}{2} = 2 x + 1$

$[∵$ Equation of the tangent at $(x_{1}, y_{1})$ is given by $T = 0$ . Here, $\frac{1}{2} (y + y_{1}) = x x_{1} + 1]$ $y = 4 x - 3$

Required area $=$ Area of shaded region

$\begin{aligned} = \int_{0}^{2} y (parabola) d x - (Area of △ P Q R) & = \int_{0}^{2} (x^{2} + 1) d x - (Area of △ P Q R) & = \frac{x^{3}}{3} + x - \frac{1}{2} 2 - \frac{3}{4} \cdot 5 \end{aligned}$

$[∵$ Area of a triangle $= \frac{1}{2} \times$ base $\times$ height $]$

$\begin{aligned} = \frac{8}{3} + 2 - 0 - \frac{1}{2} \frac{5}{4} 5 & = \frac{14}{3} - \frac{25}{8} = \frac{112 - 75}{24} = \frac{37}{24} \end{aligned}$

Area Question 10

Question 10

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Area Question 10

Question 10

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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