• question_answer The angle of intersection of the curves x2 - y2 = a2 and ${{x}^{2}}+{{y}^{2}}={{\sqrt{2a}}^{2}}$ is A)  $3\pi /4$                          B)  $\pi /2$ C)  $\pi /6$                             D)  $\pi /4$

Here, two curves are given by ${{x}^{2}}-{{y}^{2}}={{a}^{2}}$                                ...(i) and            ${{x}^{2}}+{{y}^{2}}={{\sqrt{2a}}^{2}}$                            ...(ii) From Eqs. (i) and (ii), we get ${{x}^{2}}=\frac{{{a}^{2}}}{2}(\sqrt{2}+1)$                         ?(iii) and        ${{y}^{2}}=\frac{{{a}^{2}}}{2}(\sqrt{2}-1)$                          ?(iv) Differentiating the given Eqs. (i) and (ii), we get${{m}_{1}}=\frac{dy}{dx}=\frac{x}{y},{{m}_{2}}=\frac{dy}{dx}=-\frac{x}{y}$are the slopes of the tangents at a point of intersection of the curves. If $\theta$ is the angle between the tangents $\therefore$$\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{\frac{2x}{y}}{1-\frac{{{x}^{2}}}{{{y}^{2}}}} \right|=\frac{2xy}{{{y}^{2}}-{{x}^{2}}}$ From Eqs. (iii) and (iv), we get ${{x}^{2}}{{y}^{2}}=\frac{{{a}^{4}}}{4},{{x}^{2}}-{{y}^{2}}={{a}^{2}}$                      $xy=\frac{{{a}^{2}}}{2}$ $\therefore$  $\tan \theta =\frac{{{a}^{2}}}{|-{{a}^{2}}|}=1\Rightarrow \theta =\frac{\pi }{4}$
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