AP EAPCET 24-Aug-2021 Shift 1 Paper

Given,
xy+x+y+1=0
Comparing above equation with
ax2+by2+2hxy+2gx+2fy+c=0
a=0,b=0,h=

1

2

,g=

1

2

,f=

1

2

,c=1
∵ We know that,
Product of perpendicular from origin to pair of straight lines is
|

6

√(a−b)2+4h2

|
∴P1=|

1

√(0−0)2+4( 1 2 )2

|
P1=

1

√1

=1
Now, similarly for x2−y2+2x+1=0
a=1,b=−1,h=0,g=1,f=0,c=1
Then, P2=|

1

√(1+1)2+4(0)2

|=

1

√(2)2

=

1

2

and for 2x2+3xy−2y2+2x+1=0
a=1,b=−2,h=

3

2

,g=1,f=0,c=1
Then, P3=|

1

√(1+22)+4( 3 2 )2

=

1

√9+9

=

1

3√2

∵

1

3√2

<

1

2

<1
So, p3<p2<p1