Angle between the parabolas $y^{2} = 4 (x - 1)$ and $x^{2} + 4 (y - 3) = 0$ at the common end of their latus rectum, is

{tan}^{- 1} (1)

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{tan}^{- 1} 1 + c o t^{- 1} 2 + c o t^{- 1} 3

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{tan}^{- 1} (\sqrt{3})

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{tan}^{- 1} (2) + {tan}^{- 1} (3)

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Solution

The correct option is A ${tan}^{- 1} (1)$
End points of latus rectum of $y^{2} = 4 (x - 1)$ are $(2, 2)$ and $(2, - 2)$

End points of latus rectum of $x^{2} = - 4 (y - 3)$ are $(2, 2)$ and $(- 2, 2)$

So, the common point is $(2, 2)$

$y^{2} = 4 (x - 1)$

$∴ \frac{d y}{d x} = \frac{2}{y}$

Slope of tangent at $(2, 2) = m_{1} = \frac{2}{2} = 1$

$x^{2} = - 4 (y - 4)$

$∴ \frac{d y}{d x} = \frac{x}{- 2}$

Slope of tangent at $(2, 2) = m_{2} = \frac{2}{2} = 1$

$m_{1} m_{2} = 1$

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(θ)

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