TG EAMCET 2 May 2025 Shift 1 Paper

Given equation is$\;x^2-2 x-4 y+5=0$$\Rightarrow \;\; (x-1)^2-4 y+4=0$$\Rightarrow \;\; (x-1)^2=4(y-1)$So, $(h, k)=(1,1)$ are the vertexAnd $4 a=4$$\Rightarrow a=1$Since, the parabola open upwards (because the $y$ term is positive and $x$ term is squared), the focus is $(h, k+a)=(1,1+1)=(1,2)$ and the directrix is the line $y=k-a=1-1=0$, which is the $X$-axis.Since, focal distance of a point on a parabola is its distance from the directrix.The point $(5,5)$ and directrix is $y=0$.The distance from a point $(x_0, y_0)$ to a horizontal line $y=c$ is $\left| y_0-c \right|$.$\therefore$ Focal distance $=\left|5-0\right|=5$So, the focal distance of the point $(5,5)$ on the parabola $x^2-2 x-4 y+5=0$ is 5 .