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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)
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)
y−0=3√(x−3√)y=3√x−3y2=x+2(3√x−3)2=x+23x2+9−63√x=x+23x2−x(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=(3√−x1)2+(x1+2−−−−−√)2−−−−−−−−−−−−−−−−−−−−√=3+x12−23√x1+x1+2−−−−−−−−−−−−−−−−−−−−√PB=(3√−x2)2+(x2+2−−−−−√)2−−−−−−−−−−−−−−−−−−−−√=3+x22−23√x2+x2+2−−−−−−−−−−−−−−−−−−−−√PA×PB=(5+x12−23√x1+x1)(5+x22−23√x2+x2)−−−−−−−−−−−−−−−−−−−−−−−−⎷=25+5x12+5x22−5x1(23√−1)−5x2(23√−1)+(x1x2)2−x1.x22(23√−1)−x22.x1(23√−1)+x1x2(23√−1)2−−−−−−−−−−−−−−−−−−−−−−−−−−−⎷allvaluesinaboveequationcanbecalculatedusing1and2above.calculationsarecomplex.thefinalvalueshouldbeequalto43(2+3√)Hope this information will clear your doubts about this topic.
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