Assuming that the parabola opens to the right, the form x = 1 / (4 a) * y^2 is appropriate. If it opens vertically then it's y = 1 / (4 a) * x^2.
The former form is the same as y^2 = 4 a x.
If (2, 1) is a point on a parabola with vertex at the origin you do get 1^2 = 4 a * 2 so that a = 1/8, and the equation would be y^2 = 1/2 x. Shifting the vertex to (-2, 0) the form becomes y^2 = 1/2 ( x + 2).
If the parabola is of the form y = 1 / (4a) * x^2, opening upward with vertex at the origin and containing the point (2, 1) then we would get a = 1 and the parabola would be y = 1/4 x^2. The shifted parabola would be y = 1/4 ( x + 2) ^ 2, and the focus would be at (-2, 1).
Sir,
Question# 49 Section 9.2 asks one to write an equation for the parabola with it’s vertex at (-2,0) and which has a point on the parabola of (0,1)..
Answer,
The graph is shifted in the horizontal by two units but Not in the vertical.
I know its in the form (y-k)^2=4a(x-h) however because the vertical is zero this leaves
y^2=4a(x-h)..I was struggling coming up with an answer until I decided to shift the parabola back to the origin and then solve the equation.
This meant that point (0,1) on the parobla became (2,1)
Now,
Plugging these points into y^2=4a(x-h)..I solved for (a) which equaled (1/8)
So the equation becomes y^2=(4*1/8)(x-h) which is (y-0)^2=1/2(x-(-2))…
Therefore,
If the point of the focus is (1/8,0) then y^2= 1/2 * (1/8+2)
y = +- sqrt(17)/4 and the latus rectum is (1/8,sqrt(17)/4 ) ; (1/8,-sqrt(17)/4)
I’m not sure this is correct..
Can you advise…