AP EAMCET Engineering 2011 Solved Paper

Given equation of hyperbola is
x2−3y2=3
⇒

x2

3

−

y2

1

=1
Here, a2=3 and b2=1,a>b.
Now, the equation of asymptote of this hyperbola is,
y=±

b

a

x
⇒y=±

1

√3

x
⇒ y=

x

√3

......(i)
and y=

−x

√3

.....(ii)
Let slope of asymptote (i) is, m1=

1

√3

and slope of asymptote (ii) is, m2=

−1

√3

Let θ be the angle between both asymptote, then
tan‌θ=|

m1−m2

1+m1m2

| =|

1∕√3+1∕√3

1−1∕3

|
⇒ tan‌θ=|

2∕√3

2∕3

|=√3
⇒ tan‌θ=tan‌60°
⇒ θ=π∕3