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Direct Common Tangent

The tangent a...

Question

The tangent and normal at the point $P (a t^{2}, 2 a t)$ to the parabola $y^{2} = 4 a x$ meet the x-axis in T and G respectively, then the angle at which the tangent at P to the parabola is inclined to the tangent at P to the circle through P.T.G is

{tan}^{- 1} t^{2}

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{cot}^{- 1} t^{2}

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{tan}^{- 1} t

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{cot}^{- 1} t

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Solution

The correct option is C ${tan}^{- 1} t$
Equation of tangent and normal at $P (a t^{2}, 2 a t)$ on
$y^{2} = 4 a x$ are
$t y = x + a t^{2}$ .....(1)
$y + t x = 2 a t + a t^{3}$ ...(2)
So $T (- a t^{2}, 0)$ & $G (2 a + a t^{2}, 0)$
Equation of circle passing P, T & G is
$(x + a t^{2}) (x - (2 a + a t^{2})) + (y - 0) (y - 0) = 0$
$x^{2} + y^{2} - 2 a x - a t^{2} (2 a + a t^{2}) = 0$
equation of tangent on the above circle at $P (a t^{2}, 2 a t)$ is
$a t^{2} x + 2 a t y - a (x + a t^{2}) - a t^{2} (2 a + a t^{2}) = 0$
Slope of line which is tangent to circle at $P$
$m_{1} = \frac{a (1 - t^{2})}{2 a t} = \frac{1 - t^{2}}{2 t}$
Slope of tangent at $P$ , $m_{2} = \frac{1}{t}$
$∴ tan θ = \frac{\frac{1 - t^{2}}{2 t} - \frac{1}{t}}{1 + \frac{(1 - t^{2})}{2 t^{2}}} \Rightarrow tan θ = t \Rightarrow θ = {tan}^{- 1} t$

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