The vertex of the parabola is at (1, 2) and i | vedclass

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(D) Since the axis of the parabola is parallel to the $y$-axis,its equation is of the form $(x - h)^2 = 4a(y - k)$,where $(h, k)$ is the vertex.Given

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$1/4$

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(D) Since the axis of the parabola is parallel to the $y$-axis,its equation is of the form $(x - h)^2 = 4a(y - k)$,where $(h, k)$ is the vertex.Given vertex $(h, k) = (1, 2)$,the equation becomes $(x - 1)^2 = 4a(y - 2)$.Since the parabola passes through $(0, 6)$,we substitute these coordinates into the equation:$(0 - 1)^2 = 4a(6 - 2)$$1 = 4a(4)$$1 = 16a$$a = 1/16$.The length of the latus rectum is given by $|4a|$.$|4a| = 4 	imes (1/16) = 1/4$.

The vertex of the parabola is at $(1, 2)$ and its axis is parallel to the $y$-axis. If the parabola passes through $(0, 6)$,then its latus rectum is:

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