Answer :

Given the boundaries of the area O befound are,


• Curve is (y-1)2 = 4 (x+1)


• Line y = x-1


Consider the curve


(y-1)2 = 4 (x+1)


Substitute y = x-1


(x-1-1)2 = 4(x+1)


(x-2)2 = 4x+ 4


x2 – 4x + 4 = 4x + 4


x2 – 8x = 0


x(x-8) =0


x = 8 (or) x = 0


substituting x in y = x-1


y = 7 (or) y = -1


So , the parabola meets the line y = x-1 at 2 points, D (8,7) and E (0,-1)



As per the given boundaries,


• The parabola (y-1)2 = 4 (x+1), with vertex at B(-1,1).


• Line y = x-1


The boundaries of the region to be found are,


Point B, where the curve (y-1)2 = 4 (x+1) has the extreme end the vertex i.e. B (-1,1)


Point D, where the curve (y-1)2 = 4 (x+1) and the line y= x + 1 meet i.e. D (8,7)


Point E, where the curve (y-1)2 = 4 (x+1) and the line y = x-1 meet i.e. E (0,-1)


Point O, the origin


Consider the parabola,


(y-1)2 = 4(x+1)



Area of the required region = Area of EABCDE


Area of EABCDE = Area above line ED - Area above EABCD






[Using the formula ]





The Area of the required region


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Find the area of Mathematics - Board Papers

Sketch the graph Mathematics - Board Papers

Using integrationMathematics - Board Papers

Find the area of Mathematics - Board Papers