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At the point of intersection of the rectangular hyperbola \mathrm{x y=c^2} and the parabola \mathrm{y^2=4 a x} tangents to the rectangular hyperbola and the parabola make an angle \mathrm{\theta} and \mathrm{\phi} respectively with the axis of \mathrm{\mathrm{X}}, then

Option: 1

\mathrm{\theta=\tan ^{-1}(-2 \tan \phi)}


Option: 2

\mathrm{\phi=\tan ^{-1}(-2 \tan \theta)}


Option: 3

\mathrm{\theta=\frac{1}{2} \tan ^{-1}(-\tan \phi)}


Option: 4

\mathrm{\phi=\frac{1}{2} \tan ^{-1}(-\tan \theta)}


Answers (1)

best_answer

\mathrm{\text { Let }\left(\mathrm{x}_1, \mathrm{y}_1\right) \text { be the point of intersection } \Rightarrow \mathrm{y}_1^2=4 \mathrm{ax}_1 \text { and } \mathrm{x}_1 \mathrm{y}_1=\mathrm{c}^2}

\mathrm{y^2=4 a x}                                                    \mathrm{x y=c^2}

\mathrm{\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{a}}{\mathrm{y}}}                                           \mathrm{\frac{d y}{d x}=-\frac{y}{x}}

\mathrm{\frac{\mathrm{dy}}{\mathrm{dx}}{ }_{\left(\mathrm{x}_1, \mathrm{y}_1\right)}=\tan \phi=\frac{2 \mathrm{a}}{\mathrm{y}_1}}                        \mathrm{\frac{\mathrm{dy}}{\mathrm{dx}}{ }_{\left(\mathrm{x}_1, \mathrm{y}_1\right)}=\tan \phi=\frac{y_1 }{\mathrm{x}_1}}

\mathrm{\therefore \quad \frac{\tan \theta}{\tan \phi}=\frac{-\mathrm{y}_1 / \mathrm{x}_1}{2 \mathrm{a} / \mathrm{y}_1}=\frac{-\mathrm{y}_1^2}{2 \mathrm{ax}_1}=-\frac{4 \mathrm{ax}_1}{2 \mathrm{ax}_1}=-2}

\mathrm{\Rightarrow \quad \theta=\tan ^{-1}(-2 \tan \phi)}

 

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Shailly goel

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