Find the vertex, focus, equation of axis, equation of axis, equation of directrix, length of latus rectum and the co-ordinates of the ends of latus rectum for the following parabolas :
(i) `y^(2) =8x`
(ii) `y^(2) = -12x`
(iii) `x^(2)=6y`
(iv) `x^(2)=-2y`.

Explanation

(i) `y^(2) = 8x`
Comparing with `y^(2) =4ax`
`4a = 8 rArr a=2`
`:.` Co-ordinates of vertex = (0, 0)
Co-ordinates of focus `= (alpha,0)=(2,0)`
Equation of axis `y=0`
Equation of directrix `x = -a`
`rArr x=-2`
Length of latus rectum `= 4 alpha =8`
Co-ordinates of the ends of the latus rectum
`= (a, pm 2a)`
`=(2, pm 4)`
(ii) `y^(2) = -12x`
Comparing with `y^(2)=-4ax`
`4a = 12 rArr a=3`
`:.` Co-ordiantes of vertex = (0,0)
Co-ordiantes of focus `=(-a,0)=(-3,0)`
Equation of axis `y=0`
Equation of directrix `x = a rArr x =3`
Length of latus rectum `= 4a =12`
Co-ordinates of the ends of the latus rectum
`= (-a, pm 2a)`
`=(-3, pm 6)`.
(iii) `x^(2)=6y`
Comparing `x^(2)=4ay`
`4a=6" "rArr" "a=(3)/(2)`
`:.` Co-ordinates of vertex = (0,0).
Co-ordainates of focus = `(0,a)=(0,(3)/(2))`.
Equation of axis x=0.
Equation of latus rectum y = - a
`rArr" "y=-(3)/(2)`.
Length of latus rectum = 4a=6.
Co-ordinates of the ends of the latus rectum
`(pm2a,a)=(pm3,(3)/(2))`.
(iv) `x^(2)=-2y`
Comparing with `x^(2)=-4ay`
`4a=2" "rArr" "a=(1)/(2)`
`:.` Co-ordinates of vertex = (0,0)
Co-ordinates of focus `=(0,-a)=(0,-(1)/(2))`
Equation of axis x = 0.
Equation of directrix = a.
`rArr" "y=(1)/(2)`
Length of latus rectum = 4a = 2.
Co-ordinates of the ends of latus ratus rectum
`=(pm2a,-a)=(pm1,(1)/(2))`.