Question

A normal, to parabola ${y^2} = 4ax$,whose inclination is ${30^{\circ}}$, to a parabola cuts it again at an angle of
A) ${\tan ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$
B) ${\tan ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right)$
C) ${\tan ^{ - 1}}\left( {2\sqrt 3 } \right)$
D) ${\tan ^{ - 1}}\left( {\dfrac{1}{{2\sqrt 3 }}} \right)$

Answer 1

Find the slope of tangent and normal by the equation of normal or tangent, and use the formula of angle between the normal and tangent and find the angle between them
For a parabola of equation ${y^2} = 4ax$ the tangent equation at $p(a{t_1}^2,2a{t_1})$ is ${t_1}y = x + a{t_1}^2$ where the value of the slope ${m_T} = \dfrac{1}{{{t_1}}}$

Complete step by step solution:
We know that the product of the slopes of normal and tangent is -1.
$\Rightarrow {m_T} \times {m_N} = - 1 \\\ \Rightarrow {m_N} = \dfrac{{ - 1}}{{{m_T}}} \\\ \Rightarrow {m_N} = \dfrac{{ - 1}}{{(\dfrac{1}{{{t_1}}})}} \\\ \Rightarrow {m_N} = - {t_1}.........(1) \\\$
We know that normal is always perpendicular to the tangent at point p.

We know that the equation of tangent at parabola ${y^2} = 4ax$ at $p(a{t_1}^2, 2a{t_1})$ is ${t_1}y = x + a{t_1}^2$ here, value of slope ${m_T} = \dfrac{1}{{{t_1}}}..........\left( 2 \right)$
So, the equation of normal with ${m_N} = - {t_1}$ and passing at $P(a{t_1}^2,2a{t_1})$ is

$$\Rightarrow \left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right) \\\ \Rightarrow (y - 2a{t_1}) = m(x - a{t_1}^2) \\\ \Rightarrow (y - 2a{t_1}) = - {t_1}(x - a{t_1}^2) \\\ \Rightarrow y - 2a{t_1} = - {t_1}x + a{t_1}^2 \\\ \Rightarrow y + {t_1}x = 2a{t_1} + a{t_1}^3..........\left( 3 \right) \\\$$

So, according to the question normal also cuts the curve at point $Q(a{t_2}^2,2a{t_2})$
Substitute the value of $Q(a{t_2}^2,2a{t_2})$ in the equation (3)

\Rightarrow y + {t_1}x = 2a{t_1} + a{t_1}^{3} \\\ \Rightarrow \left( {2a{t_2}} \right) + {t_1}\left( {a{t_2}^{2}} \right) = 2a{t_1} + a{t_1}^{3} \\\ \Rightarrow 2a\left( {{t_2} - {t_1}} \right) + a{t_1}({t_2}^{2} - {t_1}^{2}) = 0 \\\ \Rightarrow a\left( {{t_2} - {t_1}} \right)\left[ {{t_1}\left( {{t_1} + {t_2}} \right) + 2} \right] = 0 \\\ \Rightarrow {t_1}\left( {{t_1} + {t_2}} \right) + 2 = 0 \\\ \Rightarrow {t_2} = - {t_1} - \dfrac{2}{{{t_1}}}.........\left( 4 \right) \\\ $$ So, we know that $\tan \alpha = - {t_1}$, where $\alpha = {30^{\circ}}$, so $ - {t_1} = \dfrac{1}{{\sqrt 3 }}$ Substitute this value in equation (4), we get $ \Rightarrow {t_2} = \dfrac{{ - \left( {{t_1}^2 + 2} \right)}}{{{t_1}}} \\\ \Rightarrow {t_2} = \dfrac{{ - \left( {\dfrac{1}{3} + 2} \right)}}{{\left( {\dfrac{{ - 1}}{{\sqrt 3 }}} \right)}} \\\ \Rightarrow {t_2} = \dfrac{7}{3} \times \sqrt 3 \\\ \Rightarrow {t_2} = \dfrac{7}{{\sqrt 3 }} \\\ $ Normal cuts the curve at an angle of,

\Rightarrow \tan \Phi = \dfrac{{{m_N} - {m_T}}}{{1 + {m_N}{m_T}}} \\
\Rightarrow \tan \Phi = \dfrac{{ - {t_1} - \dfrac{1}{{{t_2}}}}}{{1 + \left( {\dfrac{{ - {t_1}}}{{{t_2}}}} \right)}} \\
\Rightarrow \tan \Phi = - \left( {\dfrac{{{t_1}{t_2} + 1}}{{{t_2}}}} \right)\left( {\dfrac{{{t_2}}}{{{t_2} - {t_1}}}} \right) \\
\Rightarrow \tan \Phi = \dfrac{{ - \left( {{t_1}{t_2} + 1} \right)}}{{{t_2} - {t_1}}} \\
\Rightarrow \tan \Phi = \dfrac{{\left( {{t_1}{t_2} + 1} \right)}}{{{t_1} - {t_2}}} \\
\Rightarrow \tan \Phi = \dfrac{{\left( {\dfrac{{ - 7}}{3} + 1} \right)}}{{ - \dfrac{1}{{\sqrt 3 }} - \dfrac{7}{{\sqrt 3 }}}} \\
\Rightarrow \tan \Phi = \dfrac{{\left( {\dfrac{7}{3} - 1} \right)}}{{\left( {\dfrac{8}{{\sqrt 3 }}} \right)}} \\
\Rightarrow \tan \Phi = \dfrac{4}{3} \times \dfrac{{\sqrt 3 }}{8} \\
\Rightarrow \tan \Phi = \dfrac{1}{{2\sqrt 3 }} \\
\Rightarrow \Phi = {\tan ^{ - 1}}\left( {\dfrac{1}{{2\sqrt 3 }}} \right) \\

So, the normal, to parabola ${y^2} = 4ax$,whose inclination is ${30^{\circ}}$, to a parabola cuts it again at an angle of $$\Phi = {\tan ^{ - 1}}\left( {\dfrac{1}{{2\sqrt 3 }}} \right)$$ **So, option D is correct.** **Note:** Always take care of the normal and tangent equations, derive the value of ${t_{1\,}}\,and\,{t_2}$, and apply in the formula of angle made by the normal when it cuts the curve. Point P and Q are two different points with different parameters.