Equation of latus rectum of parabola If parabola $\left ( y+1 \right )^{2}=k\left ( x-2 \right )$ passes through a point (1,-2), then equation of its latus rectum and directrix ?
My attempt

Is this correct answer?
because this answer is not mention in given options.
The parabola is $$(y+1)^2=(2-x)~~~~(1)$$
The most general form of parabpla is $$L_1^2=4AL_2$$ if $L_1$ and $L_2$are non-parallel. If these two are parallel and normalised $(L=\frac{ax+by+c}{\sqrt{a^2+b^2}}),$ Then length of its latus rectum is $4A$, the Eq. of directrix is $L_2=-A$, Eq. of its latus rectum is $L_2=A$. Eq. of its axis is $L_1=0$ tangent at vertex is $L_2=0$, focus is given by $(L_2=A, L_1=0)$ vertex is $(L_2=0, L_1=0).$
Eq. of directrix for (1) is $2-x=-1/4 \implies x=9/4$ and Eq. of latus rectum is $2-x=1/4 \implies x=7/4$
Vertices of parabola are (-1,2)》(h,p)
Find the k value by substituting the given point..
Now we get k=-1
General equation of parabola( Y-p)^2=-4a(X -h)
4a=k  .. a=1/4
Negative sign Denotes that parabola opens towards negative X axis
Now focus 》(2-1/4,-1)》(7/4,-1)》(h-a,p)
Which means equation of lateus rectum is x=7/4 since it passes through focus perpendicular to axis of parabola..
Now directrix passes through》(2+1/4,-1)》(9/4,-1)》(h+a,p)
Means direct passes through (9/4,-1) perpendicular to axis of parabola then its equation is x=9/4