AP EAPCET 23-Aug-2021 Shift 1 Paper

Given pair of perpendicular lines are given as, ax2+6xy+by2−10x+10y−6=0 ...(i)
General equation is given as,
ax2+2hxy+by2+2gx+2fy+c=0...(ii)
Compare Eqs. (i) and (ii), we obtain
a=a,2h=6⇒h=3, b=b, 2g=−10⇒g=−5
2f=10⇒f=5,c=−6
|

a	h	g
h	b	f
g	f	c

|=0
⇒|

a	3	−5
3	b	5
−5	5	−6

|=0
⇒a(−6b−25)−3(−18+25−5(15−5b)=0
⇒25a+25b+6ab+96=0 ...(iii)
Since, lines are perpendicular
⇒a+b=0 or a=−b ...(iv)
Use Eq. (iv) in Eq. (iii),
25(a−a)+6a(−a)+96=0
⇒6a2=96⇒a2=16
⇒a=±4
⇒|a|=4