Find the equation of parabola
(i) having its vertex at A(1,0) and focus at S(3,0)
(ii) having its focus at S(2,5) and one of the extremities of latus rectum is A (4,5)

Explanation

(i) Vertex is at A (1,0) and focus is at S (3,0).
Clearly, the axis of parabola is the x-axis.
So, we consider equation of the parabola as `(y-k)^(2)=4a(x-h)`.
Here, vertex is `(1,0)-=(h,k)`
and a = distance between vertex and focus = 2 Therefore, equation of parabola is `y^(2)=8(x-1)`.

(ii) Focus is at S (2,5) and one of the extremities of latus rectum is A (4,5).

Clearly, the other extremity of latus rectum is B(0,5) and axis of the parabola is x = 2.
So, we consider equation of parabolas as
`(x-h)^(2)=4a(y-k)`
`or(x-h)^(2)=-4a(y-k), "where "h=2`.
Length of latus rectum is AB = 4a = 4.
`:." "a=1`
Now, we have following two cases :
Parabola opens upwards :
In this case, vertex lies on the line x = 2 at distance 1 unit below the focus.
So, vertex is (2,5-1) or (2,4).
Thus, equation of parabola is `(X-2)^(2)=4(y-4)`.
Parabola this case, vertwx lies on the line x = 2 at distance 1 unit above the focus.
So, vertex is (2,5+1) or (2,6).
Thus, equation of parabola is `(X-2)^(2)=-4(y-6)`.