Given, equation of parabola is x2+10x−16y+25=0 It can be rewritten as (x+5)2=16y . . . (i) Here a=4 and vertex ≡(−5,0) and focus ≡(−5,4) So, y-coordinate of ends of latusrectum will be 4 . Putting in equation, we have ‌(x+5)2=16×4=64 ‌⇒‌‌x+5=±8 ‌ Substituting x in Eq. (i), we get y=4,4 Hence, ends of latusrectum will be (3,4) and (−13,4)