Verified Solution • 2025–26 Session moderate

QJEEM16Mad59e: JEE Main 2016 Mathematics - Class 12 JEEM16Mad59e Solution: The minimum distance of a point on the curve y = x 2 −4 from…

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

Multiple Choice Options

  • A 192{{\sqrt {19} } \over 2}
  • B 152\sqrt {{{15} \over 2}}
  • C 152{{\sqrt {15} } \over 2}
  • D 192\sqrt {{{19} \over 2}}

Step-by-Step Verified Solution

Let point on the curve

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

∴\therefore    Distance of the point ( α\alpha 2, α\alpha 2 −- 4) from origin,

D = α2+(α2−4)2\sqrt {{\alpha ^2} + {{\left( {{\alpha ^2} - 4} \right)}^2}}

⇒\Rightarrow    D2 = α\alpha 2 + α\alpha 4 + 16 −- 8 α\alpha 2

== α\alpha 4 −- 7 α\alpha 2 + 16

∴\therefore     dD2dα{{d{D^2}} \over {d\alpha }} = 4 α\alpha 3 −- 14 α\alpha

Now, dD2dα{{d{D^2}} \over {d\alpha }} = 0

⇒\Rightarrow    4 α\alpha 3 −- 14 α\alpha = 0

⇒\Rightarrow    2 α\alpha (2 α\alpha 2 −- 7) = 0

α\alpha = 0    or    α\alpha 2 = 72{7 \over 2}

d2D2dα2=12α2−14{{{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 α\alpha 2 = 72{7 \over 2}

∴\therefore    Minimum distance

D = 494−494+16\sqrt {{{49} \over 4} - {{49} \over 4} + 16}

= 152{{\sqrt {15} } \over 2}

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