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Question

If P1, P2, P3 be the product of perpendiculars from (0,0) to xy+x+y+1=0, x2−y2+2x+1=0, 2x2+3xy−2y2+3x+y+1=0 respectively then?

A
P1<P2<P3
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B
P3<P2<P1
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C
P2<P3<P1
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D
P1<P3<P2
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Solution

The correct option is B P3<P2<P1
Given
xy+x+y+1=0
comparing above eq with ax2+2hxy+by2+2gx+2fy+c=0
we get h=12,g=12,f=12,c=1
Product of perpendicular from origin is ∣ ∣c(ab)2+4h2∣ ∣
P1=∣ ∣ ∣ ∣ ∣ ∣14(12)2∣ ∣ ∣ ∣ ∣ ∣
P1=14(14)
P1=11
P1=1
Now given eq x2y2+2x+1=0
comparing above eq with ax2+2hxy+by2+2gx+2fy+c=0
we get a=1,b=1,h=0,g=1,f=0,c=1
Product of perpendicular from origin is ∣ ∣c(ab)2+4h2∣ ∣
P2=∣ ∣ ∣1(1+1)2∣ ∣ ∣
P2=14
P2=12

Now given eq 2x2+3xy2y2+3x+y+1=0
comparing above eq with ax2+2hxy+by2+2gx+2fy+c=0
we get a=2,b=2,h=32,g=32,f=12,c=1
Product of perpendicular from origin is ∣ ∣c(ab)2+4h2∣ ∣
P3=∣ ∣ ∣ ∣ ∣ ∣1(2+2)24(32)2∣ ∣ ∣ ∣ ∣ ∣
P3=∣ ∣ ∣ ∣ ∣ ∣1164(94)∣ ∣ ∣ ∣ ∣ ∣
P3=1169
P3=17

SO P3<P2<P1

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