What is the focus of the parabola y^2 - x - 2 | vedclass

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(D) Given equation: $y^2 - 2y = x - 2$.Completing the square: $(y - 1)^2 - 1 = x - 2$,which simplifies to $(y - 1)^2 = x - 1$.Let $Y = y - 1$ and $X =

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$(5/4, 1)$

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(D) Given equation: $y^2 - 2y = x - 2$.Completing the square: $(y - 1)^2 - 1 = x - 2$,which simplifies to $(y - 1)^2 = x - 1$.Let $Y = y - 1$ and $X = x - 1$. The equation becomes $Y^2 = X$.This is a standard parabola of the form $Y^2 = 4aX$,where $4a = 1$,so $a = 1/4$.The focus in the $(X, Y)$ coordinate system is $(a, 0) = (1/4, 0)$.To find the focus in the original $(x, y)$ coordinate system,we use $x = X + 1$ and $y = Y + 1$.Thus,$x = 1/4 + 1 = 5/4$ and $y = 0 + 1 = 1$.The focus is $(5/4, 1)$.

What is the focus of the parabola $y^2 - x - 2y + 2 = 0$?

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