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# if the line y-√3x+3=0 cuts the parabola y^2=x+2 at point P and Q then AP.AQ where A is(√3,0)

Soujit Dutta
19 Points
3 years ago
We may write y-√3x+3=0 as y-0/√3/2=x-√3/1/2=r....(1)Substituting x = √3+r/2, y=√3r/2 in y^2=x+2, we get 3r^2/4=r/2+√3+23r^2-2r-(4√3+8)=0AP.AQ=|r1.r2|=(4√3+8)/3=4(√3+2)/3. (Ans)
Mayur Verma
11 Points
3 years ago

y0=3(x3)y=3x3y2=x+2(3x3)2=x+23x2+963x=x+23x2x(63+1)+7=0x1+x2=63+13(1)x1.x2=73(2)NowtosolvetheproblemwetakePAandPBanypointonparabolacanbewrittenas,(x,x+2)letA(x1,x1+2)B(x2,x2+2)PA=(3x1)2+(x1+2)2=3+x1223x1+x1+2PB=(3x2)2+(x2+2)2=3+x2223x2+x2+2PA×PB=(5+x1223x1+x1)(5+x2223x2+x2)=25+5x12+5x225x1(231)5x2(231)+(x1x2)2x1.x22(231)x22.x1(231)+x1x2(231)2allvaluesinaboveequationcanbecalculatedusing1and2above.calculationsarecomplex.thefinalvalueshouldbeequalto43(2+3)