JEE Advanced 2017 Paper 1

Given: Hyperbola $\;\frac{x^{2}}{a^{2}} - \;\frac{y^{2}}{16} = 1 \;\;$ . - - - - - - - (1) Tangent $: 2x - y + 1 = 0 - \cdot s - -2$ or $2x + 1 = y \;\; \cdot s \cdot s \cdot s \cdot s (2)$ As we know, general form of tangent is $y = mx + c$ where condition of tangency $\Rightarrow c^{2} =$ $a^{2} m^{2} - b^{2} \;\; \dots (3)$ From Equation of Hyperbola, $a = a, b = 4$ And values of $c$ and $m$ from the given equation of tangent, $c = 1, m = 2$ $\Rightarrow$ In Equation $(3), \Rightarrow 1^{2} = a^{2}(2)^{2} - (4)^{2}$ $\Rightarrow 1^{2} = 4 a^{2} - 16$ $\Rightarrow 4 a^{2} = 17 \Rightarrow a = \;\frac{\pm \sqrt{17}}{2}$ Now, checking the options with Pythagoras Theorem as they are sides of right angled triangle, C) $2a, 4, 1 \Rightarrow \sqrt{17}, 4, 1 \Rightarrow \sqrt{17} > \sqrt{16} \Rightarrow \sqrt{17} > 4$ So, $\sqrt{17}$ is the bigger side. By Pythagoras theorem, $(1)^{2} + (4)^{2} = (\sqrt{17})^{2}$ $1 + 16 = 17$ It is a correct set. D) $2a, 8, 1$ $8 > \sqrt{17} \Rightarrow 8$ is bigger side. $\Rightarrow 64 = (\sqrt{17})^{2} + (1)^{2}$ $\Rightarrow 64 \neq 17 + 1$ So, they cannot be sides of the triangle. B) $a, 4, 1$ $4 > \;\frac{\sqrt{17}}{2} \Rightarrow 4$ is the biggest side. $16 = \;\frac{17}{4} + 1$ $16 \neq \;\frac{21}{4}$ They cannot be sides of a right angled triangle.A) $a, 4, 2$ $4 > \;\frac{\sqrt{17}}{2}$ $4^{2} = \left(\;\frac{\sqrt{17}}{2}\right)^{2} + (2)^{2} = \;\frac{17}{4} + 4$ $\Rightarrow 16 \neq \;\frac{33}{4}$ They cannot be sides of a right angled triangle.