Question

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

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

Answer 1

Answer: (A)

$(3,4),(-13,4)$

The equation $x^2+10x-16y+25=0$ can be written as

$(x+5{)}^2=16y$ .... $\left(i\right)$ here $a=4$ and vertex will be $-5,0$ and focus will be $-5,4$.

So, the y-coordinate of ends of latus rectum will be $4$.

Putting in equation we have,

$(x+5{)}^2=16\times 4=64$

We have $x=3,-13$

Substituting $x$ in $\left(i\right)$, we get $y=4,4$

Hence, ends of latus rectum will be $(-3,4),(-13,4)$