Q. 103.8( 73 Votes )

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

Answer :

It is given that the area of the region bounded by the parabola x^{2} = 4y and x = 4y - 2.

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

Let us calculate the point of intersection of both the curves,**Given:** x^{2} = 4y and x = 4y - 2

Therefore, putting the value of x, equation of parabola, we get,

(4y - 2)^{2} = 4y

16y^{2} + 4 - 16y = 4y

16y^{2} - 20y + 4 = 0'

4y^{2} - 5y + 1 = 0

4y^{2} - 4y - y + 1 = 0

4y(y - 1) - 1(y - 1) = 0

y = 1 or y = 1/4

Corresponding values of x are, x = 2 or x = -1

Therefore,

Coordinates of point A are

Coordinates of point B are (2, 1).

Now, draw AL and BM perpendicular to x axis.

We can see that,

Area OBCA = Area of line - Area of Parabola …(1)

Hence, the area bounded by the x^{2} = 4y and the line x = 4y – 2 is square units.

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