1.

Consider the curves C_1:y^2=4k_1x AA k_1 in [1/8,1/4] C_2:x^2=4k_2(y-2)AA k_2 in [-1,(-1)/4] C_3:x=0 Then minimum area bounded by all of these three curves is

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Solution :We need to FIND the minimum AREA BOUNDED by two paraboles (one vertical and one horizontal parabola ) with y-axis . So, the required area is minimum when horizontal parabola has MAXIMUM latus rectum (i.e. wide OPENING ) and vertical parabola has least latus rectum (i.e. narrow opening).
Hence, `k_1=1/4` and`k_2=-1/4`
So , the curves are `y^2=x` and `x^2=-(y-2)`
hence, required minimum area `underset0overset1int((2-x^2)-sqrtx)dx=1`


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