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The sum of squares of all possible values of k, for which area of the region bounded by the parabolas 2y2=kx and ky2=2(y-x) is maximum, is equal to :
8
x1= 2y2k, x2=y- ky22; let c= k2+ 2k
x1=x2 y=0 or y= 1c
Area = | 01/c (y-cy2 )dy |= | 12c2- 13c2 |= 16c2
|c|= | k2+ 2k | 2 (AM ≥ GM), area is maximum (= 124 ) when |c|=2
k2= 2k k2=4 k= 2
Sum of squares =(2)2+(-2)2=8
Answer: 8
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