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Question

Fora>0, let the curvesC1:y2=ax andC2:x2=ay intersect at originO and a point P . Let the line x=b(0<b<a) intersect the chord and the x-axis at points Q and R, respectively. If the line x=b bisects the area bounded by the curves, C1 andC2, and the area ofOQR=12, then'a' satisfies the equation.


A

x6-12x3+4=0

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B

x6-12x3-4=0

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C

x6+6x3-4=0

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D

x6-6x+4=0

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Solution

The correct option is A

x6-12x3+4=0


Explanation for the correct option:

Step 1: we will find the point of intersections of the given curves and line.

We have givenC1:y2=ax andC2:x2=ay .

Points of intersection after solving these equations areO(0,0)&P(a,a).

by joining these two point we get the chord, j

therefore, equation of chord is

y-0x-0=a-0a-0y=x

now, the point of intersection of the linex=b&x=y is(b,b).

Step 2: draw a diagram of the above information.

Step 3: Find area and solving the given things.

We have given,

area ofOQR=12

12×b×b=12b2=1b=1

Now area bounded by the two given curveC1:y2=ax andC2:x2=ay is

=0a(ax-x2a)dx=[2ax323-x33a]0a=13[2a2-a2]=a23......(1)

the linex=b divides the area bounded by the curvesC1:y2=ax andC2:x2=ay is just half of the original.

=0b(ax-x2a)dx=[2ax323-x33a]01=13[2a12-1a].......(2)

From(1)&(2),

13[2a12-1a]=12×a232a32-1=a324a32=a3+216a3=a6+4a3+4a6-12a3+4=0

Hence, option (A) is the correct option.


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