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


parabola-closed-focus

\[|PF| = \sqrt {{{\left( {x – a} \right)}^2} + {{\left( {y – 0} \right)}^2}} = \sqrt {{{\left( {x – a} \right)}^2} + {y^2}} \,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)\]

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.