Given: parabola y2=8x …(i)
And standard parabola : y2=4ax … (ii).
Compare both equation (i) and (ii), we get
a=2
Now Focus of parabola is : F(2,0)
[∵ Focus for standard parabola y2=4ax is: (a,0)]
Let P(α,β) is any point on parabola : y2=8x
Given focal distance is 4
⇒ PF=4
Distance between points (x1,y1) and (x2,y2)
=√(x2−x1)2+(y2−y1)2
⇒ Distance between F(2,0) and point P(α,β):
√(α−2)2+(β−0)2=4
Squaring on both sides
⇒(α−2)2+β2=16 …(iii)
The point P(α,β) lies on parabola y2=8x
⇒β2=8α …(iv)
Now using equation (iii) and (iv).
⇒(α−2)2+8α=16
⇒α2+4α+4=16
⇒(α+2)=±4
⇒α=−6,2
α=−6 is not possible because x coordinate of any point lying on given parabola (y2=8x) is always positive.
Putting α=2 in equation (iv)
⇒β=±4
∴ The points are (2,4) and (2,−4)
Hence, points on parabola are: (2,4) and (2,−4).