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