SATHEE: Area Question 11

Area Question 11

Question 11

The area (in sq units) of the region bounded by the curve $x^{2} = 4 y$ and the straight line $x = 4 y - 2$ is (a) $\frac{7}{8}$ (b) $\frac{9}{8}$ (c) $\frac{5}{4}$ (d) $\frac{3}{4}$

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Solution:

Given equation of curve is $x^{2} = 4 y$ , which represent a parabola with vertex $(0, 0)$ and it open upward.

Now, let us find the points of intersection of $x^{2} = 4 y$ and $4 y = x + 2$

For this consider, $x^{2} = x + 2$

$\begin{array}{lr} \Rightarrow & x^{2} - x - 2 = 0 \Rightarrow & (x - 2) (x + 1) = 0 \Rightarrow x = - 1, x = 2 \end{array}$

When $x = - 1$ , then $y = \frac{1}{4}$

and when $x = 2$ , then $y = 1$

Thus, the points of intersection are $A - 1, \frac{1}{4}$ and $B (2, 1)$ .

Now, required area $=$ area of shaded region

$\begin{aligned} = \int_{- 1}^{2} y (line) - y (parabola) d x & = \int_{- 1}^{2} \frac{x + 2}{4} - \frac{x^{2}}{4} d x = \frac{1}{4} \frac{x^{2}}{2} + 2 x - {\frac{x^{3}}{3}}_{- 1}^{2} & = \frac{1}{4} 2 + 4 - \frac{8}{3} - \frac{1}{2} - 2 + \frac{1}{3} & = \frac{1}{4} 8 - \frac{1}{2} - 3 = \frac{1}{4} 5 - \frac{1}{2} = \frac{9}{8} sq units. \end{aligned}$

Area Question 11

Question 11

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Area Question 11

Question 11

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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