Question

The ends of the latus-rectum of the conic $x^2+10x-16y+25=0$ are

• A. (3,-4), (13, 4)
• B. (-3,-4), (13,-4)
• C. (3, 4), (-13, 4)
• D. (5,-8), (-5, 8)

Answer 1

Answer: (C)

Given that, $x^2+10x-16y+25=0$
$\Rightarrow (x+5{)}^2=16y\Rightarrow X^2=4AY$
where, $X=x+5,A=4,Y=y$.
The ends of the latusrectum are $(+2A,A)$ and $(-2A,A)$
$\Rightarrow x+5=2(4)$
$\Rightarrow x=-8-5=3,y=4$
and $x+5=-2(4)$
$\Rightarrow x=-8-5=-13,y=4$
$\Rightarrow (3,4)$ and $(-13,4)$