Question

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

• A. $\left(3,-4\right),\left(13,4\right)$
• B. $\left(-3,-4\right),\left(13,-4\right)$
• C. $\left(3,-4\right),\left(-13,4\right)$
• D. $\left(5,-8\right),\left(-5,8\right)$

Answer 1

The correct option is C

$\left(3,-4\right),\left(-13,4\right)$

Explanation for correct option:

Converting the given equation to the standard equation of a parabola

$$\begin{gathered} x^2+10x-16y+25=0 \\ \Rightarrow {\left(x+5\right)}^2=16y \\ \Rightarrow X^2=4AY \end{gathered}$$

where, $X=x+5,A=4,Y=y$

Finding the ends of the latus rectum

The ends of the latus rectum are $\left(2A,A\right),\left(-2A,A\right)$with respect to the standard equation.

This gives,

$$\begin{gathered} x+5=8 \\ \Rightarrow x=3,y=4 \end{gathered}$$

And

$$\begin{gathered} x+5=-8 \\ \Rightarrow x=-13,y=4 \end{gathered}$$

So, the end of the latus rectum are $\left(3,-4\right),\left(-13,4\right)$

Hence, the correct answer is Option (C).