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

JEE · Math · previous-year question

  1. A.$${{\sqrt {19} } \over 2}$$
  2. B.$$\sqrt {{{15} \over 2}} $$
  3. C.$${{\sqrt {15} } \over 2}$$correct
  4. D.$$\sqrt {{{19} \over 2}} $$

Answer

C. $${{\sqrt {15} } \over 2}$$

Explanation

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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