## EXERCISE 2.1

#### Page No 365:

#### Question 1:

Find the area of the region bounded by the curve *y*^{2} = *x* and the lines *x* = 1, *x* = 4 and the *x*-axis.

#### Answer:

The area of the region bounded by the curve, *y*^{2} = *x*, the lines,* x* = 1 and* x* = 4, and the *x*-axis is the area ABCD.

#### Question 2:

Find the area of the region bounded by *y*^{2} = 9*x*,* x* = 2, *x* = 4 and the *x*-axis in the first quadrant.

#### Answer:

The area of the region bounded by the curve, *y*^{2} = 9*x*, *x* = 2, and* x* = 4, and the *x*-axis is the area ABCD.

#### Page No 366:

#### Question 3:

Find the area of the region bounded by *x*^{2} = 4*y*, *y* = 2, *y* = 4 and the *y*-axis in the first quadrant.

#### Answer:

The area of the region bounded by the curve, *x*^{2} = 4*y*, *y* = 2, and* y* = 4, and the *y*-axis is the area ABCD.

#### Question 4:

Find the area of the region bounded by the ellipse

#### Answer:

The given equation of the ellipse, , can be represented as

It can be observed that the ellipse is symmetrical about *x*-axis and *y*-axis.

∴ Area bounded by ellipse = 4 × Area of OAB

Therefore, area bounded by the ellipse = 4 × 3π = 12π units

#### Question 5:

Find the area of the region bounded by the ellipse

#### Answer:

The given equation of the ellipse can be represented as

It can be observed that the ellipse is symmetrical about *x*-axis and *y*-axis.

∴ Area bounded by ellipse = 4 × Area OAB

Therefore, area bounded by the ellipse =

#### Question 6:

Find the area of the region in the first quadrant enclosed by *x*-axis, line and the circle

#### Answer:

The area of the region bounded by the circle, , and the *x*-axis is the area OAB.

The point of intersection of the line and the circle in the first quadrant is .

Area OAB = Area ΔOCA + Area ACB

Area of OAC

Area of ABC

Therefore, required area enclosed =

32 + π3 – 32 = π3 square units

#### Question 7:

Find the area of the smaller part of the circle *x*^{2} +* y*^{2} = *a*^{2} cut off by the line

#### Answer:

The area of the smaller part of the circle, *x*^{2} +* y*^{2} = *a*^{2}, cut off by the line, , is the area ABCDA.

It can be observed that the area ABCD is symmetrical about *x*-axis.

∴ Area ABCD = 2 × Area ABC

Therefore, the area of smaller part of the circle, *x*^{2} +* y*^{2} = *a*^{2}, cut off by the line, , is units.

#### Question 8:

The area between *x* = *y*^{2} and *x* = 4 is divided into two equal parts by the line *x* = *a*, find the value of *a*.

#### Answer:

The line, *x* = *a*, divides the area bounded by the parabola and *x* = 4 into two equal parts.

âˆ´ Area OAD = Area ABCD

It can be observed that the given area is symmetrical about *x*-axis.

â‡’ Area OED = Area EFCD

From (1) and (2), we obtain

Therefore, the value of *a* is .

#### Question 9:

Find the area of the region bounded by the parabola *y *= *x*^{2} and

#### Answer:

The area bounded by the parabola, *x*^{2} = *y*,and the line,, can be represented as

The given area is symmetrical about *y*-axis.

âˆ´ Area OACO = Area ODBO

The point of intersection of parabola, *x*^{2} = *y*, and line, *y *= *x*, is A (1, 1).

Area of OACO = Area Î”OAM â€“ Area OMACO

Area of Î”OAM

Area of OMACO

â‡’ Area of OACO = Area of Î”OAM â€“ Area of OMACO

Therefore, required area = units

#### Question 10:

Find the area bounded by the curve *x*^{2} = 4*y* and the line *x* = 4*y *– 2

#### Answer:

The area bounded by the curve, *x*^{2} = 4*y*, and line, *x* = 4*y *– 2, is represented by the shaded area OBAO.

Let A and B be the points of intersection of the line and parabola.

Coordinates of point .

Coordinates of point B are (2, 1).

We draw AL and BM perpendicular to *x*-axis.

It can be observed that,

Area OBAO = Area OBCO + Area OACO … (1)

Then, Area OBCO = Area OMBC – Area OMBO

Similarly, Area OACO = Area OLAC – Area OLAO

Therefore, required area =

#### Question 11:

Find the area of the region bounded by the curve *y*^{2} = 4*x* and the line *x* = 3

#### Answer:

The region bounded by the parabola, *y*^{2} = 4*x*, and the line, *x* = 3, is the area OACO.

The area OACO is symmetrical about *x*-axis.

∴ Area of OACO = 2 (Area of OAB)

Therefore, the required area is units.

#### Question 12:

Area lying in the first quadrant and bounded by the circle *x*^{2} + *y*^{2} = 4 and the lines *x* = 0 and *x *= 2 is

**A.** π

**B.**

**C.**

**D. **

#### Answer:

The area bounded by the circle and the lines, *x* = 0 and *x* = 2, in the first quadrant is represented as

Thus, the correct answer is A.

#### Question 13:

Area of the region bounded by the curve *y*^{2} = 4*x*, *y*-axis and the line *y* = 3 is

**A.** 2

**B.**

**C.**

**D. **

#### Answer:

The area bounded by the curve, *y*^{2} = 4*x*, *y*-axis, and *y* = 3 is represented as

Thus, the correct answer is B.