AP EAMCET Engineering 22 Sep 2020 Shift 1 Paper

Let the lines represented by equation
ax2+2hxy+by2+2gx+2fy+c=0
arc y=m1x+c1 and y=m2x+c2
So, ax2+2hxy+by2+2gx+2fy+c
=(m1x−y+c1)(m1x−y+c2)
⇒‌‌‌

m1m2

a

=‌

−(m1+m2)

2h

=‌

1

b

Now, according to the information given in question, as lines are equidistance from origin, so
‌

|C1|

√1+m12

=‌

|C2|

√1+m22

⇒‌‌C12(1+m22)=C22(1+m12)
⇒‌‌C12−C22=C22m12−C12m22
From Eq. (i), we have
C1+C2=−‌

2f

b

and m1C2+m2C1=‌

2g

b

and C1C2=‌

c

b

and m1m2=‌

9

b

and Therefore from Eq. (ii), we have
⇒f2(f2−bc)=g2(g2−ac) (on squaring both sides)
⇒f4−g4=c(bf2−ag2)