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  • If a normal to a parabola ax make an angle <img alt="\mathrm{\phi}" src="https://entrance

If a normal to a parabola \mathrm{y^2=4} ax make an angle \mathrm{\phi} with its axis, then it will cut the curve again at an angle

Option: 1

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


Option: 2

\mathrm{\tan ^{-1}\left(\frac{1}{2} \tan \phi\right)}


Option: 3

\mathrm{\cot ^{-1}\left(\frac{1}{2} \tan \phi\right)}


Option: 4

\mathrm{\text { none }}


Answers (1)

best_answer

\mathrm{\text { normal at } t: \quad y+t x=2 a t+a t^2}

\mathrm{\therefore \quad \mathrm{m}_{\mathrm{N}} \text { at } \mathrm{A}=-\mathrm{t}=\tan \phi}

\mathrm{\mathrm{t}=-\tan \phi=\mathrm{m}_1}

\mathrm{\text { Now tangent at B } \quad t_1 y=x t+a t_1^2}

\mathrm{\mathrm{m}_{\mathrm{T}} \text { at } \mathrm{A}=\frac{1}{\mathrm{t}_1}=\mathrm{m}_2}                       \mathrm{\text { also } t_1=-t-\frac{2}{t}}

\mathrm{\therefore \quad \tan \theta=\left|\frac{\frac{1}{t_1}+\mathrm{t}}{1-\frac{t}{t_1}}\right|=\left|\frac{1+t_1}{t-t_1}\right|=\left|\frac{1+t^2}{2\left(t+\frac{1}{t}\right)}\right|=\left|\frac{\sec ^2 \phi \cdot \tan \phi}{2\left(\sec ^2 \phi\right)}\right| \quad\left[\text { As } \mathrm{t} \mathrm{t}_1=-\mathrm{t}^2-2\right]}

\mathrm{\text { Hence } \tan \theta=\frac{\tan \phi}{2} \Rightarrow \theta=\tan ^{-1}\left(\frac{\tan \phi}{2}\right)}

Posted by

Gautam harsolia

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