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Question

# If the line y−√3x+3=0 cuts the parabola y2=x+2 at A and B, then PA.PB is equal to [where P≡(√3,0)]

A
4(3+2)3
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B
4(23)3
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C
4(3)2
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D
2(3+2)3
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Solution

## The correct option is A 4(√3+2)3Consider the equation of the line. y=√3x−3 Consider the equation of the parabola. y2=x+2 And, ⇒P(√3,0) Find the intersection points. (√3x−3)2=x+2 3x2+9−6√3x=x+2 3x2−x(6√3+1)+7=0 ⇒x1+x2=6√3+13 ......(1) ⇒x1x2=73 ......(2) Any point the parabola can be written as, ⇒(x,√x+2) Let, A=(x1,√x1+2) B=(x2,√x2+2) Therefore, PA=√(√3−x1)2+(√x1+2)2 ⇒PA=√3+x21−2√3x1+x1+2 PB=√(√3−x2)2+(√x2+2)2 ⇒PB=√3+x22−2√3x2+x2+2 Therefore, PA×PB=√(5+x21−(2√3−1)x1)(5+x22−(2√3−1)x2) PA×PB=43(√3+2)=4(√3+2)3 Hence, this is the required result.

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