A) \[3\pi /4\]
B) \[\pi /2\]
C) \[\pi /6\]
D) \[\pi /4\]
Correct Answer: D
Solution :
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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