Question

The parabolas `y^2=4xa n dx^2=4y` divide the square region bounded by the lines `x=4,y=4` and the coordinate axes. If `S_1,S_2,S_3` are the areas of these parts numbered from top to bottom, respectively, then `S_1: S_2-=1:1` (b) `S_2: S_3-=1:2` `S_1: S_3-=1:1` (d) `S_1:(S_1+S_2)=1:2`
A. `1:1:1`
B. `2:1:2`
C. `1:2:3`
D. `1:3:2`

Answer 1

Correct Answer - a
Clearly, `y^(2)=4xand x^(2)=4y` intersect at (4,4).
`therefore S_(2)+S_(3)=underset(0)overset(4)(int)sqrt(4x)dx`
`impliesS_(2)+S_(3)=2underset(0)overset(4)(int)sqrtxdx=4/3[x^(3//2)]_(0)^(4)=32/3` sq. units
and `S_(3)=underset(0)overset(4)(int)(x^(2))/(4)dy=1/4[(x^(3))/(3)]_(0)^(4)=16/3` sq. units

Thus, we have
`S_(2)+S_(3)=32/3and S_(3)=16/3impliesS_(2)=16/3` sq. units.
`becauseS_(1)+S_(2)+S_(3)="Area of square"OAPB=4xx4=16` sq. units.
`therefore S_(2)=16/3 "sq. units." " "[because S_(2)=S_(3)=16/3"sq. units"]`
Hence, `S_(1):S_(2):S_(3)=1:1:1`