find the vertex, focus, diretrix and length of latus-rectum of the parabola

- y
^{2}- 4y - 2x - 8 = 0 - 2y
^{2}+ 3y - 4x -3 = 0 - 9y
^{2}- 16x - 12y - 57 = 0

the given equation of parabola is:

shifting the origin to the point (-6,2) without rotating the axes and denoting the new coordinates with respect to these new axes by X and Y, we have

x=X-6 , y=Y+2 .....(2)

using these relations , equation (1) reduces to .....(3)

this is of the form . on comparing we get 4a=2 ie. a=1/2

**vertex**: the coordinates of the vertex with respect to the new axes are X=0, Y=0

so, the coordinates of the vertex with respect to the old axes are (-6,2).

**focus**: the coordinates of the focus with respect to the new axes are (X=a, Y=0) i.e. (X=1/2, Y=0)

so, the coordinates of the focus with respect to the old axes are

**directrix**: the equation of the directrix with respect to the new axes is X=-a i.e.

so, the equation of the directrix with respect to the old axes is

**latus rectum**: the length of the latus-rectum of the parabola is equal to 4a=2

i hope you can solve (ii) and (iii) question in the similar way.

**
**