Q.

Let m1 and m2 be the slopes of the tangents drawn from the point P(4, 1) to the hyperbola H: y225-x216=1. If Q is the point from which the tangents drawn to H have slopes |m1| and |m2| and they make positive intercepts α and β on the x-axis, then (PQ)2αβ is equal to _________ .        [2023]


Ans.

(8)

Equation of tangent to the hyperbola y225-x216=1

y=mx±25-16m2                                    ...(i)

Equation (i) passes through (4,1),

1=4m±25-16m2⇒1-4m=±25-16m2

⇒1+16m2-8m=25-16m2⇒32m2-8m-24=0

⇒4m2-m-3=0⇒4m2-4m+3m-3=0

⇒4m(m-1)+3(m-1)=0

⇒(4m+3)(m-1)=0⇒m=1,-34

It is given that |m1| and |m2| are the slopes of tangents, so

|m1|=1 and |m2|=34

Equations of tangents with positive slope are:

y=x±25-16 ⇒y=x±3

and y=34x±25-16×916⇒4y=3x±16

For positive intercept on the x-axis, y=x-3 and 4y=3x-16

∴  α=163 and β=3

Solve y=x-3 and 4y=3x-16 to find their intersection point.

⇒ x=-4 and y=-7

So, Q≡(-4,-7)

(PQ)2=(4+4)2+(1+7)2=64+64=128

∴  (PQ)2αβ=12816=8