The minimum distance of a point on the curve y = x2-4 from the origin is :

Question

Accepted Answer

Let point on the curve

y = x2 - 4 is ($$\alpha $$2, $$\alpha $$2 - 4)

$$ \therefore $$   Distance of the point ($$\alpha $$2, $$\alpha $$2 - 4) from origin,

D = $$\sqrt {{\alpha ^2} + {{\left( {{\alpha ^2} - 4} \right)}^2}} $$

$$ \Rightarrow $$   D2 = $$\alpha $$2 + $$\alpha $$4 + 16 - 8$$\alpha $$2

$=$ $$\alpha $$4 - 7$$\alpha $$2 + 16

$$ \therefore $$    $${{d{D^2}} \over {d\alpha }}$$ = 4$$\alpha $$3 - 14$$\alpha $$

Now, $${{d{D^2}} \over {d\alpha }}$$ = 0

$$ \Rightarrow $$   4$$\alpha $$3 - 14$$\alpha $$ = 0

$$ \Rightarrow $$   2$$\alpha $$ (2$$\alpha $$2 - 7) = 0

$$\alpha $$ = 0    or   $$\alpha $$2 = $${7 \over 2}$$

$${{{d^2}{D^2}} \over {d{\alpha ^2}}} = 12{\alpha ^2} - 14$$

$$ \therefore $$   $${\left( {{{{d^2}{D^2}} \over {d{\alpha ^2}}}} \right)_{at\,\,\alpha  = 0}} =  - 14 < 0$$

$${\left( {{{{d^2}{D^2}} \over {d{\alpha ^2}}}} \right)_{at\,\,{\alpha ^2} = {7 \over 2}}} = 28 > 0$$

$$\therefore\,\,\,$$ Distance is minimum at $$\alpha $$2 = $${7 \over 2}$$

$$ \therefore $$   Minimum distance

D = $$\sqrt {{{49} \over 4} - {{49} \over 4} + 16} $$

= $${{\sqrt {15} } \over 2}$$

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