Question 7 - Jee advanced Math 2022 P2 Questions with Solutions

 Consider the hyperbola

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

with foci at \(S\) and \(S_{1}\),  where \(S\) lies on positive x-axis. Let \(P\) be a point on the hyperbola, in the first quadrant. Let \(\angle{SPS_{1}} = \alpha\), with \(\alpha < \frac{\pi}{2}\). 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 \(S_{1}P\) at \(P_{1}\). Let \(\delta\) be the distance of \(P\) from the straight line \(SP_{1}\), and \(\beta = S_{1}P\). Then the greatest integer less than or equal to \(\frac{\beta \delta}{9}\sin{\frac{\alpha}{2}}\) is ______.

Sol : 

There are a couple of properties of hyperbola that will help us here.

(1) The tangent line at a point bisects the angle between the lines connecting the two foci with the point of tangency. 

So, in the figure above, tangent at \(P\) bisects the angle \(\alpha\) between the lines joining the foci(\(S\) and \(S_{1}\)) with the point of tangency \(P\). 

\(\implies \angle{SPT} = \angle{S_{1}PT} = \frac{\alpha}{2}\) where \(T\) is the point where the tangent meets the x-axis. 

(2) The second property of hyperbola required here states that the product of the lengths of the segments from the foci onto a tangent is equal to \(b^{2}\). 

Dropping perpendicular segments from \(S\) and \(S_{1}\) onto the tangent at \(P\). Let \(R\) and \(R’\) be the points where they meet the tangent.

\(\implies SR \times S_{1}R’ = b^{2}\)

Since the lines \(R’P\) and \(SP_{1}\) are parallel, \(SR = \delta\)

In right angled triangle \(S_{1}R’P\),

\(\sin{\frac{\alpha}{2}} = \frac{S_{1}R’}{\beta}\)

\(\implies S_{1}R’ = \beta \sin{\frac{\alpha}{2}}\)

Comparing the given equation of the hyperbola \(\frac{x^{2}}{100} - \frac{y^{2}}{64} = 1\) with the standard form \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\).

\(\implies b^{2} = 64\)

\(SR \times S_{1}R’ = b^{2}\)

\(\implies \delta \times \beta \sin{\frac{\alpha}{2}} = 64\)

\(\implies \frac{\beta \delta}{9}\sin{\frac{\alpha}{2}} = \frac{64}{9}\)

\(7 < \frac{64}{9} < 8\)

Therefore, the greatest integer less than or equal to \(\frac{\beta \delta}{9}\sin{\frac{\alpha}{2}}\) is \(7\).



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