Q. The ends of the latusrectum 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) $(5, - 8),\ ( - 5,\ 8)$
Explanation
(c)
Given, $(x + 5)^{2} = 16y$
$\Rightarrow X^{2} = 4AY\ \ where\ X = x + 5,\ A = 4,\ Y = y.\$
The ends of the latusrectum are
$(2A,\ A\ )$ and $( - 2A,A)$
$\Rightarrow x + 5 = 2(4),y = 4 \Rightarrow \ x = 3,y = 4$
and $x + 5 = - 2(4),\ y = 4 \Rightarrow \ x = - 13,y = 4$
Here required points are (3, 4) and $( - 13,\ 4)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.