TS EAMCET 20-July-2022 Shift 2 Solved Paper

Curve I x2−y2=4
On differentiating w.r.t. x, we get‌2x−2yy′=0
⇒‌‌x=yy′
⇒‌‌y′=‌

x

y

....(i)
⇒‌‌m1=‌

x

y

Curve II y2=3x
On differentiating w.r.t. x, we get
‌2yy′‌=3×1
⇒‌y′‌=‌

3

2y

⇒‌m2‌=‌

3

2y

.....(ii)
To get intersection point,
x2−4=y2,y2=3x
On putting x2−4=3x,
‌x2−3x−4=0
‌⇒‌‌x2−4x+x−4=0
‌⇒‌‌x(x−4)+1(x−4)=0‌. ‌
‌⇒‌‌(x−4)(x+1)=0
‌x=4,−1
From the graph it is clear that x=4∴‌‌y2=3×4
⇒‌‌y=±2√3
At point (4,2√3) from Eqs. (i) and (ii), we get
m1‌=‌

4

2√3

=‌

2

√3

m2‌=‌

3

2(2√3)

=‌

√3

4

tan‌θ‌=|‌

m1−m2

1+m1m2

|