Question

The extremities of the latus rectum of the parabola are $(7,5)$ and $(7,3)$. Find its equation.

• A. ${(y-4)}^2=2(x-\frac{13}{2})$
• B. ${(y-4)}^2=-2(x-\frac{13}{2})$
• C. ${(y-4)}^2=-2(x-\frac{15}{2})$
• D. ${(y-4)}^2=2(x-\frac{15}{2})$

Answer 1

Answer: (A), (C)

${(y-4)}^2=2(x-\frac{13}{2})$
${(y-4)}^2=-2(x-\frac{15}{2})$

Equation of L.R. is
$x=7,$
Also focus is mid - point of L.R. is $(7,4).$
$\therefore$ Equation of axis is $y-4=0(x-7,),m=0$ as it is
$\perp$ to L.R. $\therefore y=4$ is the equation of axis and
length of L.R. $=2=4a,\therefore a=\frac{1}{2}$
Hence its vertex is at a distance $a=\frac{1}{2}$ from focus $S$
on the axis $y=4.$
$\therefore$ Vertex is $(7\pm \frac{1}{2},4)$ or $(7\frac{1}{2},4)$ and $(6\frac{1}{2},4)$
Hence its equation is $Y^2=4AX$ or $Y^2=-4AX$
where $Y=y-4,X=x-6.5$ or $x-7.5$ and $4A=2$
${(y-4)}^2=2(x-6.5)$ or ${(y-4)}^2=-2(x-7.5)$
Ans: A,C