The focal distance of a point (5, 5) on the p | vedclass

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Question

(A) Given the equation of the parabola: $x^2 - 2x - 4y + 5 = 0$. Rearranging the terms to complete the square: $x^2 - 2x + 1 = 4y - 5 + 1$. $(x - 1)^2

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$5$

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(A) Given the equation of the parabola: $x^2 - 2x - 4y + 5 = 0$. Rearranging the terms to complete the square: $x^2 - 2x + 1 = 4y - 5 + 1$. $(x - 1)^2 = 4y - 4$. $(x - 1)^2 = 4(y - 1)$. Comparing this with the standard form $(x - h)^2 = 4a(y - k)$,we get $h = 1$,$k = 1$,and $4a = 4$,so $a = 1$. The focal distance of a point $(x_1, y_1)$ on a parabola $(x - h)^2 = 4a(y - k)$ is given by $|y_1 - k + a|$. Substituting the point $(5, 5)$ and the values $k = 1, a = 1$: Focal distance $= |5 - 1 + 1| = |5| = 5$.

The focal distance of a point $(5, 5)$ on the parabola $x^2 - 2x - 4y + 5 = 0$ is

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