AP EAPCET 05-Jul-2022 Shift 2 Paper

Let y=mx be a common line to both
2x2+ax(mx)+3m2x2=0
and 2x2+bx(mx)−3m2x2=0
Common from x2 in both equation,
and 2+ma+3m2‌=0
2+mb−3m2‌=0
Adding both equation, we get
‌m(a+b)=−4
⇒‌‌m=‌

−4

a+b

Subtracting both equations, we get
m(a−b)+6m2=0
‌⇒m≠0,‌ so ‌m=‌

−(a−b)

6

‌⇒‌‌‌

−4

a+b

=‌

−(a−b)

6

‌⇒a2−b2=24
Now, let y=m1x and y=‌

−1

m1

x be two perpendicular line.
Solving, we get m1=‌

−13

2a+3b

and m1=−‌

(3a−2b)

13

⇒‌(2a+3b)(3a−2b)‌=169
⇒6a2−4ab+9ab−6b2‌=169
⇒‌6(24)+5ab‌=169
⇒‌5ab‌=25⇒ab=5
⇒‌a2−(‌

5

a

)2‌=24⇒a2−‌

25

a2

=24
‌⇒‌‌a4−24a2−25=0
‌⇒‌‌a2=25 [∵a2≠−1]
‌⇒‌‌a=5‌ and ‌b=1