`y^2+2y-x+5=0` represents a parabola. Find its vertex, equation of axis, equation of latus rectum, coordinates of the focus, equation of the directrix, extremities of the latus rectum, and the length of the latus rectum.

Explanation

We have `y^(@)+2y-x+5=0`
`or" "x=y^(2)+2y+5` (1)
`:." "(y+1)^(2)=(x-4)` (2)
This is parabola having vertex at A(4,-1) and axis parallel to x-axis having equation y=1=0.

Equation of tengent at vertex is x-4=0.
Also, coefficient of `y^(2)` in (1) is positive.
So, parabola opens to the right.
Comparing (2) with `(y-k)^(2)=4a(x-h)`, we have `4a=1ora=(1)/(4)`.
Thus, length of latus rectum is 1 unit.
Focus is on the axis at distance a units from the vertex.
`:." "Focus,S-=(4+(1)/(4),-1)-=((17)/(4),-1)`
Also, equation of directrix is `x=4-(!)/(4)orx=(15)/(4)`.
Ends point of latus rectum lies on either side of focus at distance 2a units on the katus rectum line.
Thus, end points of latus rectum are `((17)/(4),-1pm(1)/(2))-=((17)/(4),-(1)/(2))and((17)/(4),-(3)/(2))`.