Find the focus and directrix of $y^2=4x+1$ I am teaching myself maths by following a textbook. I have just started a section on parabolas. I read that the equation $y^2=4ax$ represents a parabola with focus (a,0), directrix $x=-a$. I also understand how this is derived from the parametric equations $x=at^2, y=2at$. Then come introductory, simple exercises. But I fall at the first hurdle. I am asked to sketch parabolas , showing their foci and directrices. The first ones I can do. E.g. for $y^2=4x$ it is easy to see that $a=1$, so the focus is (1,0) and the directrix -1. But what about $y^2=4x+1$? I can see from graphing software that the parabola shifts to the left by $\frac{1}{4}$ but cannot see how this relates to $y^2=4ax$. I have consulted textbooks and the internet but cannot come to grips with this concept.
 A: You can notice see that
$$y^2 = 4x+1 = 4(x+\frac{1}{4})$$
Let $x'  = x + \frac{1}{4}$
Your new equation is
$$y^2 = 4x'$$
Hence, the focus in the coordinates $(x',y)$ is $(1, 0)$ which in coordinates $(x,y)$ will be $(\frac{3}{4},0)$
A: You can represent $y^2=4x+1$ in the following way:
$$y^2=4x+1$$
$$y^2=4(x+\frac 14)$$
Now, let us consider $x+\frac 14=X$ and $y=Y$. The geometric significance of this substitution is that, now we have shifted our origin to $(-\frac 14,0)$. So we get the standard form $Y^2=4(1)X$ when we consider origin to be $(-\frac 14,0)$. Hence, with respect to this shifted frame, focus would be at $X=1$ which means that $x+\frac 14=1$ so $x=\frac 34$ with respect to the frame in the original question. Now try to find the equation of directrix in a similar manner.
A: Yes, the parabola $y^2 = 4ax$ or $x^2 = 4 a y$ is the simplest form. But parabola can be shifted or rotated or both shifted and rotated. Rotated would mean, its axis of symmetry is neither horizontal nor vertical.
For example,
$(y - k)^2  =  4a(x - h)$ is general equation of a shifted horizontal parabola with vertex at $(h, k)$.
In this particular case,
$(y-0)^2 = 4x+1 = 4 \left(x + \dfrac{1}{4}\right)$
So, $h = - \dfrac{1}{4}, k = 0, a = 1$
Hence directrix is $x = - a + h = - \dfrac{5}{4}$ and the same shift applies to focus too.
A: General form of a horizontal  parabola :
$$\tag{1} y^2=4 px  $$
Question :
$$y^2 = 4x+1  $$
Can be expressed as :
$$  y^2 = 4 \left (x + \frac{1}{4} \right)$$
In general form :
$$ \tag{2} (y-0)^2 = 4\left( x - \left( -\frac{1}{4} \right) \right)$$
Note the vertex of this parabola is at :
$$(h,k) = \left(-\frac{1}{4}, 0 \right)$$
Note :
Comparing two equation (1) and (2) we know :
$$ p = 1$$
Focus coordinates : $$( h+p ~,\, k) => \left( \frac{3}{4}, 0\right) $$
Directrix equation :
$$ x = h-p => x = -\frac{5}{4}$$

Note : the first one seemed easy because that was considering the vertex of the parabola to be the origin $(0,0)$, but in the Question asked it’s $\left(-\frac{1}{4}, ~0\right)$ so a bit of algebra required.
