Hello, Shivanand!
Find the vertex and latus rectum of the parabola: $\displaystyle (y+3)^2 \:=\:2(x+2)$
To find the vertex n latus rectum is easy through standard formula
but the given formula is not in standard form . . . . It isn't? We are given: . $\displaystyle (y+3)^2 \:=\:2(x+2)$
We already know:
. . that the vertex is at (-2,-3)
. . that the parabola opens to the right
. . that $\displaystyle p = \tfrac{1}{2}$
So the parabola looks like this: Code:
L *
*
* :
* : 2p
* :
:
V * - - - * F - - - - - -
p :
* :
* : 2p
* :
*
R *
The latus rectum is $\displaystyle LR.$
I don't see any difficulty . . .