The angle between the parabolas y^2 = 4(x-1) | vedclass

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(A) For the parabola $y^2 = 4(x-1)$,the vertex is $(1, 0)$ and $a = 1$. The latus rectum is $x = 1+1 = 2$. The ends of the latus rectum are $(2, 2)$ a

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$\frac{\pi}{2}$

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(A) For the parabola $y^2 = 4(x-1)$,the vertex is $(1, 0)$ and $a = 1$. The latus rectum is $x = 1+1 = 2$. The ends of the latus rectum are $(2, 2)$ and $(2, -2)$.For the parabola $x^2 = -4(y-3)$,the vertex is $(0, 3)$ and $a = 1$. The latus rectum is $y = 3-1 = 2$. The ends of the latus rectum are $(2, 2)$ and $(-2, 2)$.The common point is $(2, 2)$.For $y^2 = 4(x-1)$,differentiating with respect to $x$: $2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}$. At $(2, 2)$,$m_1 = \frac{2}{2} = 1$.For $x^2 = -4(y-3)$,differentiating with respect to $x$: $2x = -4 \frac{dy}{dx} \implies \frac{dy}{dx} = -\frac{x}{2}$. At $(2, 2)$,$m_2 = -\frac{2}{2} = -1$.Since $m_1 	imes m_2 = 1 	imes (-1) = -1$,the tangents are perpendicular.Therefore,the angle between the parabolas is $\frac{\pi}{2}$.

The 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:

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