Question

The area bounded by the curve $y ^ { 2 } = 4 x$ and $x ^ { 2 } = 4 y$ is

• A. $\frac { 16 } { 3 }$ sq. units
• B. $\frac { 3 } { 16 }$sq. units
• C. $\frac { 14 } { 3 }$ sq. units
• D. $\frac { 3 } { 14 }$sq. units

Answer 1

Answer: (A)

$\frac { 16 } { 3 }$ sq. units

Answer 2

Required area $= \int _ { 0 } ^ { 4 } ( O A B C - O D B C )$ Region

$= \int _ { 0 } ^ { 4 } \left( \sqrt { 4 x } - \frac { x ^ { 2 } } { 4 } \right)$ dx = $\frac { 16 } { 3 }$ square unit.

Trick : From Important Tips’ the area of the region bounded by $y ^ { 2 } = 4 a x$ and $x ^ { 2 } = 4 b y$ is $\frac { 16 a b } { 3 }$ square unit.

Here $y ^ { 2 } = 4 x$ and $x ^ { 2 } = 4 y$ so a = 1 and b = 1

Required area = square unit.

$D _ { i }$