Let y=x+1 is axis of parabola, y+x-4=0 is tangent of same parabola at its vertex and y=2x+3 is one of its tangents. Then find the focus of the parabola.

Answer

`((17)/(9),(26)/(9))`

Explanation

We have
axis : `L-=x-y+1=0`
Tangent at vertex : `T_(1)-=x+y-4=0`
Tangent at some point on the parabola : `T-=2x-y+3=0`

Let the focus be S(h,k).
Since focus lies on the axis L,
h-k1=0 (1)
Now, foot of perpendicular from focus upon tangent T lies on tangent at vertex `T_(1)`.
So, sloving tangents `T_(1)andT`, we get M `((1)/(3),(11)/(3))`, which is required foot of perpendicular.
Now, slope of T is 2.
`:." Slope of SM"=(k-(11)/(3))/(h-(1)/(3))=-(1)/(2)`
`or" "3h+6k-23=0` (2)
Solving (1) and (2), we get
`h=(17)/(9)andk=(26)/(9)`
`:." Focus"-=((17)/(9),(26)/(9))`