Question

Find the equation of the parabola where extremities of latus rectum are $(3,5)$ and $(3,-3).$

• A. ${(y-1)}^2=-8(x-5)$
• B. ${(y-1)}^2=-4(x-5)$
• C. ${\left(y\right)}^2=-8(x-5)$
• D. ${(y-1)}^2=-8(2x-5)$

Answer 1

The correct option is A ${(y-1)}^2=-8(x-5)$

Clearly $L.R$. is parallel to $y-$ axis therefore axis is parallel to $x-$ axis.
$L.R.=4a=\pm 8.$ If its vertex be $(h,k),$ then
${(y-k)}^2=8(x-h)$
Make it pass through the given points and solving
for $(h,k),$ we get $h=k=1$
$\therefore {(y-1)}^2=8(x-1).$
Similarly ${(y-1)}^2=-8(x-5)$