Question

Prove that the point on the parabola y² = 4ax(a > 0) nearest to focus is its vertex.

Answer 1

Let point (x, y) on parabola y² = 4ax is nearest point to focus of parabola.

\(\therefore\) distance between (x, y) & focus(a, 0)

\(\leq a\) (\(\because\) a is distance between focus & vertex)

⇒ (x - a)² + (y - 0)² \(\leq a^2\)

⇒ y² + (x - a)² \(\leq a^2\) 

⇒ 4ax + (x - a)² \(\leq a^2\) (\(\because\) y² = 4ax)

⇒ (x + a)² \(\leq a^2\)

⇒ |x + a| \(\leq a\)

⇒ -a \(\leq \) x + a \(\leq \) a

⇒ - 2a \(\leq \) x \(\leq \) 0

But x \(\geq\) 0 for given parabola

\(\therefore\) x = 0 is only possibility

If x = 0 then y = 0

\(\therefore\) (0,0) is nearest point to focus which is vertex of the parabola.

Hence proved.