MHT CET202616 April 2026Evening ShiftMathematicsArea Under CurvesActual
The area of the region common to the parabolas 4y^2 = 9x and 3x^2 = 16y is...
Options
- A2 sq. units
- B4 sq. units
- C8 sq. units
- D16 sq. units
Correct answer
B. 4 sq. units
Step-by-step solution
The given equations of the parabolas are 4y^2 = 9x and 3x^2 = 16y . From the second equation, we have y = 3x^2 16 . Substituting this into the first equation: 4 ( 3x^2 16 )^2 = 9x 9x^4 64 = 9x x^4 - 64x = 0 x(x^3 - 64) = 0 The real roots are x = 0 and x = 4 . The required area A is the integral of the upper curve minus the lower curve from x = 0 to x = 4 : A = ₀⁴ ( 3 2 x - 3x^2 16 ) dx A = [ x^ 3/2 - x^3 16 ]₀⁴ A = ( 4^ 3/2 - 4^3 16 ) - 0 A = 8 - 4 = 4 sq. units. Answer: 4 sq. units