TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

We have, y2=4x
⇒‌‌2y‌

dy

dx

=4⇒‌

dy

dx

=‌

2

y

At A(1,2)
‌

dy

dx

|(1,2)=‌

2

2

=1
∴ Tangent is given by
y−2=1(x−1)
y−2=x−1
y=x+1......(i)
At B(4,−4)
(‌

dy

dx

)(4,−4)=‌

2

−4

=−‌

1

2

∴ Tangent is given by y+4=‌

−1

2

(x−4)
2y+8=−x+4
x+2y+4=0.......(ii)
At C(2,2√2)
‌

dy

dx

|(2,2√2)=‌

2

2√2

=‌

1

√2

∴ Tangent is given by
y−2√2=‌

1

√2

(x−2)
⇒‌‌√2y−4=x−2
⇒‌‌x−√2y+2=0..........(iii)
Let P,Q,R the vertices of triangle formed by tangents (i), (ii), (iii). On solving Eqs. (i), (ii) and (iii), we get
P(−2,−1),Q(−2√2,√2−2) and R(√2,√2+1)
Now, ar (∆ABC)=‌

1

2

|

1	2	1
4	−4	1
2	2√2	1

|
and ar =‌

1

2

[6−3√2−6]=‌

−3

2

√2
=‌

3

2

√2sq [area cannot be negative]
∴‌‌α=3√2 and β=‌

3

2

√2
∴‌‌αβ=3√2×‌

3

2

√2=9