find the area of the region included between the parabola y^2 = 3x^2/4 n the line 3x-2y+12=0 - Maths - Application of Integrals

Question

find the area of the region included between the parabola y^2 =
3x^2/4 n the line 3x-2y+12=0

Ajanta Trivedi · Accepted Answer

if the given parabola is y =
3(x^2)/4 (1) and the equation of the line is
3x-2y+12=0 (2)
eq(1) represents the parabola with
vertex (0,0) and axis of the parabola is along the positive
direction of y-axis
eq(2) represents the straight line
which makes x = -12/3 = -4 intercept with x-axis
and y=12/2 = 6 ,i e line intersect
the y-axis at
(0,6)

the point of intersection of (1) and
(2) is
x= 4 and x = -2
therefore y=12 or
y=3
points are (4, 12) and (-2,
3)
Required area
=∫-24y2 dx-∫-24y1 dx=∫-243x+122 dx-∫-2434 x2 dx=123
x22+12x|-24-34 x33|-24=1232*16+12*4-32*4-12*(-2)-14*43-(-2)3=12
24+48-6+24-14*64+8=12*90-14*72=45-18=27 sq units

hope this helps you