area common to y2=ax and x2+y2=4ax specific doubt: i am getting points of intersection of 2 curves as x=0 and - Maths - Application of Integrals

Question

area common to y2=ax and
x2+y2=4ax
specific doubt: i am getting points of intersection of 2 curves as
x=0 and x=3a , it is looking wrong

Manbar Singh · Accepted Answer

The given equations are :y2 = ax 1x2 + y2 = 4ax 2Eq1 represents a parabola having the vertex at the origin and axis along x-axis Eq2 represents a circle having the centre at 2a, 0 and radius 2a POINTS OF INTERSECTION :Put y2 = ax in 2, we get x2 + ax = 4ax⇒x2 - 3ax = 0⇒xx - 3a = 0⇒x = 0 or x-3a = 0⇒x = 0 or x = 3aWhen x = 0; y = 0When x = 3a; y = ±3aSo, points of intersection are : 0, 0 and 3a, ±3aSo, P3a, 3a; Q3a, -3a and O0,0 are the points of intersection

area common to y2=ax and x2+y2=4ax specific doubt: i am getting points of intersection of 2 curves as x=0 and x=3a , it is looking wrong

area common to y²=ax and x²+y²=4ax
specific doubt: i am getting points of intersection of 2 curves as x=0 and x=3a , it is looking wrong