Intersection point of {y}^{2}=8x&y=\sqrt{2}x is (0,0)&(2,4\sqrt{2})

Area of ΔABC=21(2)⋅1=21
So required area=area between the curve y2=8x and the line y=2x−ar△ABC
=[∫04(8x−2x)dx]−21
=3322−82−21=6132
JEE Main 2022 — Mathematics Calculus
The area enclosed by y2=8x and y=2x that lies outside the triangle formed by y=2x,x=1,y=22, is equal to
Held on 29 Jun 2022 · Verified 6 Jul 2026.
6162
6112
6132
652
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