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The tangent and normal at $\mathrm{\mathrm{P}(\mathrm{t}),}$ for all real positive $\mathrm{ \mathrm{t},}$ to the parabola $\mathrm{\mathrm{y}^2=4 \mathrm{ax}}$ meet the axis of the parabola in T and G respectively, then the angle at which the tangent at P to the parabola is inclined to the tangent at $\mathrm{\mathrm{P}}$ to the circle passing through the points $\mathrm{\mathrm{P}, \mathrm{T}}$ and $\mathrm{\mathrm{G}}$ is
Option: 1
$\mathrm{\cot ^{-1} t}$

Option: 2
$\mathrm{\cot ^{-1} t^2}$

Option: 3
$\mathrm{\tan ^{-1} t}$

Option: 4
$\mathrm{\tan ^{-1} t^2}$

Answers (1)

best_answer

$\mathrm{\text { slope of tangent }=\frac{1}{t}\left(\mathrm{~m}_1\right) \text { at } \mathrm{P} \text { on parabola }}$

$\mathrm{\text { slope of } P S=\frac{2 a t}{a\left(t^2-1\right)}=\frac{2 t}{t^2-1}}$

$\mathrm{\therefore \quad \text { slope of tangent at } P \text { on circle }=\frac{1-t^2}{2 t}\left(\mathrm{~m}_2\right)}$

$\mathrm{\therefore \quad \tan \theta=\frac{\frac{1}{\mathrm{t}}-\frac{1-\mathrm{t}^2}{2 \mathrm{t}}}{1+\frac{1-\mathrm{t}^2}{2 \mathrm{t}^2}}=\frac{\left(2-1+\mathrm{t}^2\right) 2 \mathrm{t}^2}{2 \mathrm{t}\left(1+\mathrm{t}^2\right)}=\mathrm{t}}$

$\mathrm{\therefore \quad \theta=\tan ^{-1} \mathrm{t} \Rightarrow(\mathrm{C})}$

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