Given equation is ‌x2−2x−4y+5=0 ⇒‌‌(x−1)2−4y+4=0 ⇒‌‌(x−1)2=4(y−1) So, (h,k)=(1,1) are the vertex And 4a=4 ⇒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 (x0,y0) to a horizontal line y=c is |y0−c|. ∴ Focal distance =|5−0|=5 So, the focal distance of the point (5,5) on the parabola x2−2x−4y+5=0 is 5 .