The set of values of α2, if there exists a tangent to the ellipse x2α2+y2=1 such that the portion of the tangent intercepted by the hyperbola α2x2−y2=1 subtends a right angle at the centre of the curves, is
A
[√5+12,2]
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B
[1,2]
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C
[√5−12,1]
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D
[√5−12,√5+12]
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Solution
The correct option is A[√5+12,2] Equation of any tangent to x2α2+y2=1 in slope form is, y=mx±√α2m2+1 ⇒(y−mx√α2m2+1)2=1
By homogenisation, we can write (α2x2−y2)(α2m2+1)=y2+m2x2−2mxy
The portion of tangent subtends 90∘ at the origin. coefficient ofx2+coefficient ofy2=0 ⇒(α2m2+1)(α2−1)=1+m2 ⇒α2(α2−1)m2+α2−1=1+m2 ⇒[α2(α2−1)−1]m2=2−α2 ⇒m2=2−α2α4−α2−1≥0 ⇒α2−2α4−α2−1≤0