JEE Main & Advanced Sample Paper JEE Main Sample Paper-5

  • 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)  \[\pi /4\]

    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 \[{{x}^{2}}{{y}^{2}}=\frac{{{a}^{4}}}{4},{{x}^{2}}-{{y}^{2}}={{a}^{2}}\]                      \[xy=\frac{{{a}^{2}}}{2}\] \[\therefore \]  \[\tan \theta =\frac{{{a}^{2}}}{|-{{a}^{2}}|}=1\Rightarrow \theta =\frac{\pi }{4}\]

You need to login to perform this action.
You will be redirected in 3 sec spinner