Question

Find the equation of the parabola if the focus is at (a, 0) and the vertex is at (a', 0) 

Answer 1

In a parabola, the vertex is the mid-point of the focus and the point of intersection of the axis and the directrix.

Let (x₁, y₁) be the coordinates of the point of intersection of the axis and directrix. 

 It is given that the vertex and the focus of a parabola are (a', 0) and (a, 0), respectively.

Thus, the slope of the axis of the parabola is zero.

And, the slope of the directrix cannot be defined.

Let the directrix intersect the axis at K (r, s). 

∴ $$\frac{r+a}{2}=a^{\prime },\frac{s+0}{2}=0$$
$$\Rightarrow r=2a^{\prime }-a,s=0$$ 

∴ Required equation of the directrix is $$x-2a^{\prime }+a=0$$ 

Let P (x, y) be any point on the parabola whose focus is S (a, 0), and the directrix is $$x-2a^{\prime }+a=0$$ 

Draw PM perpendicular to $$x-2a^{\prime }+a=0$$ 

Then, we have: 

$$SP=PM$$
$$\Rightarrow S{P}^2=P{M}^2$$
$$\Rightarrow {(x-a)}^2+{(y-0)}^2={\left(\frac{x-2a^{\prime }+a}{\sqrt{1}}\right)}^2$$
$$\Rightarrow y^2={(x-2a^{\prime }+a)}^2-{(x-a)}^2$$
$$\Rightarrow y^2=x^2+4a^{\prime 2}+a^2-4a^{\prime }x-4a{a}^{\prime }+2ax-x^2-a^2+2ax$$
$$\Rightarrow y^2=4a^{\prime 2}-4a^{\prime }x-4a{a}^{\prime }+4ax$$
$$\Rightarrow y^2=-4(a^{\prime }-a)(x-a^{\prime })$$