# The Point of Parabola Closed to Focus is the Vertex

The point of parabola closed to focus is the vertex. Let the given parabola be
${y^2} = 4ax\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

The vertex and the focus of the parabola are $O\left( {0,0} \right)$ and $F\left( {a,0} \right)$ respectively, as shown in the given diagram. Let $P\left( {x,y} \right)$ be any point on the parabola, then its distance from the focus is $|PF| = \sqrt {{{\left( {x – a} \right)}^2} + {{\left( {y – 0} \right)}^2}} = \sqrt {{{\left( {x – a} \right)}^2} + {y^2}} \,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

Putting the value of ${y^2}$ from equation (i) and equation (ii), we have

$|PF| = \sqrt {{{\left( {x – a} \right)}^2} + 4ax} = \sqrt {{{\left( {x + a} \right)}^2}} = x + a\,\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right)$

The distance of the vertex from the focus is

$|OF| = a\,\,\,\,{\text{ – – – }}\left( {{\text{iv}}} \right)$

It is also clear from the above diagram that for any point $P\left( {x,y} \right)$ of the parabola,
$\begin{gathered} x > 0 \Rightarrow x + a > a \\ \Rightarrow |PF| > |OF| \\ |OF| > |PF| \\ \end{gathered}$

This shows that the distance of the vertex from the focus is less than the distance of $P$ from the focus, so the point of the parabola closed to focus is the vertex.