Question

Find the area common to the circle x^2+y^2=16a^2 and the parabola y^2=6ax

Answer 1

This question done by concept of circle and parabola... And application of integration....

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Answer 2

• Eq. of circile witrh center at the origin and radius 4a it is given as             $x^{2}$+$y^{2}$=16$a^{2}$ ----------(1)

• Eq. of parabola with vertex at the origin and axis along x-axis is               $y^{2}$=6ax ----------(2)

• To find the points of intersection of the given curves, we solve these eq. simultaneously.

         substituting y from eq. (2) in eq. (1)

          $x^{2}$+6ax-16$a^{2}$=0

          (x+8a)(x-2a)=0

          x=2a,-8a

         now putting vaule of x in eq. 2

        at x=2a        y=$\sqrt{12a^{2} }$

        y=±2$\sqrt{3}$a

• So the two curves intersect at (2a,2$\sqrt{3}$a) and (2a,-2$\sqrt{3}$a)
• Clearly, both the curves are symmetrical about x- axis.
• So, Required area=2(area of OAB)