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Q.

The area (in square units) of the region described by

$(x, y) y 2 ≤ 2 x a n d y ≥ 4 x − 1 i s :$

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a

732

b

564

c

1564

d

932

answer is D.

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Detailed Solution

Here, we have to find the area between the parabola

y 2 ≤ 2 x

and the line

y ≥ 4 x − 1

.
Now

y 2 ≤ 2 x

will give the inside part of the parabola and

y ≥ 4 x − 1

. Will give the left side part of the line in the graph. Here, we need to find the points of intersection of the curves to determine the value of the limits required for integration. So, let us solve the given equations by considering the equality sign. Substituting

y = 4 x − 1

in

y 2 = 2 x

we get,

⇒ 16 x 2 + 1 − 8 x = 2 x ⇒ 16 x 2 − 10 x + 1 = 0

Now, using the discriminant method, we get,

⇒ x = − (− 10) ± (− 10) 2 − 4 × 1 × 16 2 × 16

⇒ x = 10 ± 100 − 64 32

⇒ x = 10 ± 6 32

⇒ x = 1632; x = 432

⇒ x = 12; x = 18

When

x = 12

⇒ y = 4 × 12 − 1 ⇒ y = 2 − 1 ⇒ y = 1

When

x = 18

⇒ y = 4 × 18 − 1 ⇒ y = 12 − 1 ⇒ y = − 1 2

Hence, the coordinates of points of intersection can be given as

A 12, 1

and

B 18, − 1 2

. Let us plot the graph of the given situation. So, the required graph will look like:

https://www.vedantu.com/question-sets/dd9e0067-4035-4f54-987e-15ff12d10c832615872494846346200.png

Clearly, we can see that the area of the region bounded by the curves is shown by the shaded part. Therefore, using integration, we have,

⇒ A r e a = ∫ a b [f (y) − g (y)] d y

Here

f (y)

is the equation of the line in terms of

y

.

= y + 1 4

And,

g (y)

is the equation of the parabola in terms of

y

.

= y 2 2

Here, ‘a’ and ‘b’ are the limits up to which the area between the graph is to be determined.

⇒ (a, b) = − 1 2, 1 ⇒ Area = ∫ − 1 1 y + 1 4 − y 2 2 d y ⇒ Area = 14 y 2 2 + y − y 3 2 × 3 − 1 2 1

Substituting the limits and simplifying, we get,

⇒ Area = 38 − 16 − − 3 32 + 148 ⇒ Area = 1048 + 332 − 148 ⇒ Area = 932

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