The correct option is A [√5+12,2]
Equation of tangent at point P(acosθ,sinθ) is
xacosθ+y1sin θ=1⋯(1)
Let, it cut the hyperbola at points P and Q
Homogenizing the hyperbola a2x2−y2=1 with the help of the above the equation, we get
a2x2−y2=(xacosθ+ysinθ)2
This is a pair of straight lines OP and OQ
∵OP⊥OQ
⇒Coefficient of x2+coefficient of y2=0
⇒a2−cos2θa2−1−sin2θ=0
⇒a2−cos2θa2−1−1+cos2θ=0
⇒cos2θ=a2(2−a2)a2−1
Now, 0≤cos2θ≤1
⇒0≤a2(2−a2)a2−1≤ 1
Let t=a2, t>0
Case (i) 0≤t(2−t)(t−1)⇒ t(t−2)(t−1)≤0
⇒ t∈(1,2]⋯(1) (∵ t>0)
Case (ii) t(2−t)(t−1)≤1
⇒ −t2+t+1t−1≤0
⇒ (t−1+√52)(t−1−√52)(t−1)≥0
⇒ t∈[0,1)∪[1+√52,∞)⋯(2) (∵ t>0)
Hence required solution is
∴a2∈[√5+12,2]