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