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Question

The exhaustive set of values of a2 such that there exists a tangent to the ellipse x2+a2y2=a2 such that the portion of the tangent intercepted by the hyperbola a2x2y2=1 subtends a right angle at the origin.

A
[5+12,2]
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B
(1,2]
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C
[512,1)
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D
[512,1)(1,5+12]
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Solution

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 a2x2y2=1 with the help of the above the equation, we get
a2x2y2=(xacosθ+ysinθ)2
This is a pair of straight lines OP and OQ
OPOQ
Coefficient of x2+coefficient of y2=0
a2cos2θa21sin2θ=0
a2cos2θa211+cos2θ=0
cos2θ=a2(2a2)a21
Now, 0cos2θ1
0a2(2a2)a21 1
Let t=a2, t>0
Case (i) 0t(2t)(t1) t(t2)(t1)0
t(1,2](1) ( t>0)
Case (ii) t(2t)(t1)1
t2+t+1t10
(t1+52)(t152)(t1)0
t[0,1)[1+52,)(2) ( t>0)
Hence required solution is
a2[5+12,2]

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