Find the equation of each of the following parabola
focus at `(-1,2)` directrix `x-2y+3=0`

Explanation

(i) Given that , directrix=0 and focus =(6,0)
So, the equation of the parabola
`(x-6)^(2)+y^(2)=x^(2)`
`rArr x^(2)+36-12x+y^(2)=x^(2)` ltbr. `rArr y^(2)-12x+36=0`
(ii) given that, vertex=(0,4) and focus=(0,2)

So, the equation of parabola is
`sqrt((x-0)^(2)+(y-2)^(2))=abs(y-6)`
`rArr x^(2)+y^(2)-4y+4=y^(2)-12y+36`
`rArr x^(2)-4y+12y-32=0`
`rArr x^(2)+8y-32=0`
`rArr x^(2)=32-8y`
(iii) Given that, focus at (-1,-2) and directrix x-2y+3=0
So, the equation of paarobola is `sqrt((x-1)^(2)+(y-2)^(2))=abs((x-2y+3)/(sqrt(1+4)))`
`rArr x^(2)+2x+1+y^(2)+4y+4=1/5[x^(2)+4y^(2)+9-4xy-12y+6x]`
`rArr 4x^(2)+4xy+y^(2)+4x+32y+16 =0`