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Question

The point P(26,3) lies on the hyperbola x2a2y2b2=1 having eccentricity 52. If the tangent and normal at P to the hyperbola intersect its conjugate axis at the points Q and R respectively, then QR is equal to

A
36
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B
6
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C
63
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D
43
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Solution

The correct option is C 63
P(26,3) lies on the hyperbola x2a2y2b2=1
24a23b2=1 (i)
Now, b2a2=e21=14 [e=52]
4b2=a2 (ii)
From (i) and (ii), we get
a2=12 and b2=3
Hyperbola: x212y23=1
Tangent at P(26,3),
x6y3=1Q(0,3)
Slope of tangent =12
Normal at P(26,3),
y3=2(x+26)
R(0,53)
QR=63

Alternate Solution:
Let P(asecθ, btanθ)
T:xasecθybtanθ=1 Q(0, btanθ)
N:axsecθ+bytanθ=a2+b2R(0,a2+b2btanθ)
QR=(a2+b2b)tanθ+btanθ
b2a2=e21=144b2=a2
a2=12 and b2=3

secθ=2 and tanθ=1

Now, QR=∣ ∣(12+33)1+31∣ ∣=63

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