- Home
- Standard 11
- Mathematics
At the point of intersection of the rectangular hyperbola $ xy = c^2 $ and the parabola $y^2 = 4ax$ tangents to the rectangular hyperbola and the parabola make an angle $ \theta $ and $ \phi $ respectively with the axis of $X$, then
$\theta = tan^{-1} (- 2 tan\phi )$
$\phi = tan^{-1}(- 2 tan\theta )$
$\theta = tan^{-1} (- tan\phi )$
$\phi = tan^{-1} (- tan\theta )$
Solution
Let $ (x_1, y_1)$ be the point of intersection $ \Rightarrow $ $y_1^2\, = \,4a{x_1}\,$ and $ x_1y_1 = c^2$
$Y^2 = 4ax $ $xy = c^2$
$\frac{{dy}}{{dx}}\, = \,\frac{{2a}}{y}\,$ $\frac{{dy}}{{dx}}\, = \, – \,\,\frac{y}{x}\,\,$
${\frac{{dy}}{{dx}}_{({x_1},{y_1})}}\, = \,\,\,\tan \phi \,\,\, = \,\,\,\frac{{2a}}{{{y_1}}}\,$ ${\frac{{dy}}{{dx}}_{({x_1},{y_1})}}\, = \,\tan \theta \, = \, – \,\frac{{{y_1}}}{{{x_1}}}$
$\frac{{\tan \theta }}{{\tan \phi }}\,\, = \,\,\frac{{ – {y_1}/{x_1}}}{{2a/{y_1}}}\,\, = \,\,\frac{{ – y_1^2}}{{2a{x_1}}}\,\, = \,\, – \,\,\frac{{4a{x_1}}}{{2a{x_1}}}\,\, = \,\, – 2$
$\Rightarrow$ $\theta = \tan^{-1} (- 2 \tan\phi )$
