Equation of latus rectum of parabola

Question

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.

Answer 1

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 $2-x=1/4 \implies x=7/4$

Answer 2

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