Hyperbola

Given equation of hyperbola can be written as
‌

x2

5

−‌

y2

4

=1which is of the form
‌

x2

a2

−‌

y2

b2

=1‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
where,a=√5,b=2
As we know that, equation of any tangent to the hyperbola Eq. (i) having slopemis
y=mx±√a2m2−b2=mx+√5m2−4‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
According to question, Eq. (ii) passes through(1,1).
So,1=m×1±√5m2−4
‌⇒‌‌1−m=±√5m2−4
‌⇒‌‌(1−m)2=5m2−4
‌⇒‌‌4m2+2m−5=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(iii)
Letm1,m2slopes of these two tangents be the roots of Eq. (iii).
So,m1+m2=‌

−2

4

=‌

−1

2

‌m1m2=‌

−5

4

‌∴tan‌θ=|‌

m1−m1

1+m1m2

|
‌‌=|‌

√(m1+m2)2−4m1m2

1+m1m2

|
‌‌=|‌

√( −1 2 )2−4×( −5 4 )

1+(‌ −5 4 )

|
‌‌=|‌

√ 1 4 + 20 4

1−‌ 5 4

|=|‌

√21

2

×‌

4

−1

|
‌‌=|−2√21|=2√21