The ends of the latusrectum of the parabola $x^{2}+10 x-16 y+25=0$ are

Given, equation of parabola is $$ x^{2}+10 x-16 y+25=0 $$ It can be rewritten as $$ (x+5)^{2}=16 y $$ Here $a=4$ and vertex $\equiv(-5,0)$ and focus $=(-5,4)$ So, $y$-coordinate of ends of latusrectum will be 4 . Putting in equation, we have $(x+5)^{2}=16 \times 4=64$ $\Rightarrow \quad x+5=\pm 8$ So, $\quad x=3,-13$ Substituting $x$ in Eq. (i), we get $y=4,4$ Hence, ends of latusrectum will be $(3,4)$ and $(-13,4)$.