`y^2+2y-x+5=0` represents a parabola. Find equation of 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))`.