TS EAMCET 9 May 2024 Shift 1 Solved Paper

We have, y2=8x⇒a=2
Any tangent to parabola is
y=mx+‌

a

m

⇒mx−y+‌

2

m

=0
If it is a tangent to the circle x2+y2=2, then perpendicular from centre (0,0) is equal to radius √2.
‌∴‌

| 2 m |

√m2+1

=√2⇒‌

4

m2

=2(m2+1)
‌⇒m4+m2−2=0
‌⇒(m2+1)(m2−1)=0⇒m=±1
Hence, the common tangents are.
‌y=±(x+2)
‌x+y+2=0‌ or ‌x−y+2=0
Since, −‌

a

b

>0, so in this case,
equation of tangent : x−y+2=0
On comparing with ax+by+2=0, we get a=1,b=−1
So, 3a2+2b+1=3−2+1=2