$Let a and b be positive real numbers such that a > 1 and b < a . Let P be a point in the first quadrant$
$that lies on the hyperbola \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 . Suppose the tangent to the hyperbola at P passes through the$
$point (1, 0), and suppose the normal to the hyperbola at P cuts off equal intercepts on the coordinate$
$axes. Let Δ denote the area of the triangle formed by the tangent at P, the normal at P and the$
$x -axis. If e denotes the eccentricity of the hyperbola, then which of the following statements is/are TRUE?$

Difficult

Solution

Since Normal at point P makes equal intercept on co-ordinate axes, therefore slope of Normal = –1
Hence slope of tangent = 1
Equation of tangent
$\begin{array}{l} y - 0 = 1 (x - 1) \\ y = x - 1 \\ Equation of tangent at (x_{1} y_{1}) \end{array}$
$\begin{array}{l} \frac{{xx}_{1}}{a^{2}} - \frac{{yy}_{1}}{b^{2}} = 1 \\ x - y = 1 (equation of Tangent) \\ on comparing x_{1} = a^{2}, y_{1} = b^{2} \end{array}$
$\begin{array}{l} {Also a}^{2} - b^{2} = 1 - - - (1) \\ Now equation of normal at (x_{1} y_{1}) = ({a^{2}}_{1} b^{2}) \end{array}$
$\begin{array}{l} y - b^{2} = - 1 (x - a^{2}) \\ x + y = a^{2} + b^{2} \dots (Normal) \\ point of intersection with x -axis is (a^{2} + b^{2}) \end{array}$
$\begin{array}{l} Now e = \sqrt{1 + \frac{b^{2}}{a^{2}}} \\ e = \sqrt{1 + \frac{b^{2}}{b^{2} + 1}} [from (1) \frac{b^{2}}{b^{2} + 1} < 1] \\ 1 < e < \sqrt{2} option (A) \\ Δ = \frac{1}{2} \cdot AB \cdot PQ \\ and Δ = \frac{1}{2} (a^{2} + b^{2} - 1) \cdot b^{2} \\ Δ = \frac{1}{2} (2 b^{2}) b^{2} (from (1) a^{2} - 1 = b^{2}) \\ Δ = b^{4} so option (D) \end{array}$

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