QJEEAdv20Ma9eb4: JEE Advanced 2020 Mathematics - Class 11 JEEAdv20Ma9eb4 Solution: Let a and b be positive real numbers such that a > 1 and b < a. …

Equation of given hyperbola is

${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$ , a > b and a > 1

Let point P(a sec $\theta$ , b tan $\theta$ ) on the hyperbola in first quadrant i.e., $Qe\left( {0,{\pi \over 2}} \right)$ .

Now equation of tangent to the hyperbola at point P is

${{x\sec \theta } \over a} - {{y\tan \theta } \over b} = 1$ .... (i)

$\because$ The tangent (i) passes through point A(1, 0)

So, a = \sec \theta &gt; 1 . ....(ii)

and equation of normal to the hyperbola at point P having slope is ${{b\sec \theta } \over {a\tan \theta }}$ , as normal cuts off equal intercepts on the coordinate axes, so slope must be $-$ 1.

Therefore, ${{b\sec \theta } \over {a\tan \theta }}$ = $-$ 1

$\Rightarrow$ $b = - \tan \theta$ {from Eq. (ii)}

$\because$ The eccentricity of hyperbola

$e = \sqrt {1 + {{{b^2}} \over {{a^2}}}} = \sqrt {1 + {{\sin }^2}\theta }$

$\therefore$ 1 &lt; e &lt; \sqrt 2 , as $\theta \in \left( {0,{\pi \over 2}} \right)$

And the area of required triangle is, which is isosceles is

$\Delta = {1 \over 2}{(AP)^2} = {1 \over 2}[{(1 - {\sec ^2}\theta )^2} + {\tan ^4}\theta ]$



$= {1 \over 2}(2{\tan ^4}\theta ) = {\tan ^4}\theta = {b^4}$ .