A line x+1=y meets the curve 2x3+10x2+x−4=y at A,B and C. If point P≡(−1,0), then |PA.PB.PC| is equal to
A
3√2
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B
3
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C
3√3
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D
2√3
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Solution
The correct option is A3√2 Parametric form of the line y=x+1 is ycos45°=x+1sin45°=r⇒x=r√2−1,y=r√2
Putting the values of x,y in the equation of the curve
we get 2(r√2−1)3+10(r√2−1)2+(r√2−1)−4=(r√2)⇒r3√2+2r2−7√2r+3=0...(1)
If the roots of the equation are r1,r2&r3∴r1r2r3=−3√2∴|PA.PB.PC|=3√2