JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Hyperbola

Consider the hyperbola

\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1

with foci at S and

S_{1}

, where S lies on the positive x-axis. Let P be a point on the hyperbola, in the first quadrant. Let

∠ S P S_{1} = α

, with

α < \frac{π}{2}

. The straight line passing through the point S and having the same slope as that of the tangent at P to the hyperbola, intersects the straight line

S_{1} P

P_{1}

. Let

δ

be the distance of P from the straight line

S P_{1}

, and

β = S_{1} P

. Then the greatest integer less than or equal to

\frac{β δ}{9} sin \frac{α}{2}

is ____.

Answer: 7

Solution

For the hyperbola

\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1

, we have

a = 10

b = 8

, and

c = 100 + 64 = 164 = 241

.

The foci are at

S = (241, 0)

and

S_{1} = (- 241, 0)

.

From the hyperbola definition:

S_{1} P - S P = 2 a = 20

, so

β - \frac{δ}{s i n \frac{α}{2}} = 20

.

Using the cosine rule in triangle

S P S_{1}

cos α = \frac{S P ^{2} + β ^{2} - 4 c ^{2}}{2 β \cdot S P}

With

S P = \frac{δ}{s i n \frac{α}{2}}

and

4 c^{2} = 656

, after algebraic manipulation:

\frac{β δ}{sin \frac{α}{2}} \cdot 2 sin^{2} \frac{α}{2} = 128

\frac{β δ}{9} sin \frac{α}{2} = \frac{64}{9} \approx 7.11

Therefore,

[\frac{β δ}{9} sin \frac{α}{2}] = 7

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Integer answer