QJEEM19Ma1729: JEE Main 2019 Mathematics - Class 12 JEEM19Ma1729 Solution: A helicopter is flying along the curve given by y – x 3/2 = 7,…

Question

Verified Solution • 2025–26 Session moderate

A helicopter is flying along the curve given by y – x^3/2 = 7, (x

\ge

0). A soldier positioned at the point

\left( {{1 \over 2},7} \right)

wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is -

Accepted Answer

y - {x^{3/2}} = 7\left( {x \ge 0} \right)

{{dy} \over {dx}} = {3 \over 2}{x^{1/2}}

\left( {{3 \over 2}\sqrt x } \right)\left( {{{7 - y} \over {{1 \over 2} - x}}} \right) = - 1

\left( {{3 \over 2}\sqrt x } \right)\left( {{{ - {x^{3/2}}} \over {{1 \over 2} - x}}} \right) = - 1

{3 \over 2}.{x^2} = {1 \over 2} - x

3{x^2} = 1 - 2x

3{x^2} + 2x - 1 = 0

3{x^2} + 3x - x - 1 = 0

\left( {x + 1} \right)\left( {3x - 1} \right) = 0

\therefore

x = - 1

(rejected)

x = {1 \over 3}

y = 7 + {x^{3/2}} = 7 + {\left( {{1 \over 3}} \right)^{3/2}}

{\ell _{AB}} = \sqrt {{{\left( {{1 \over 2} - {1 \over 3}} \right)}^2} + {{\left( {{1 \over 3}} \right)}^3}} = \sqrt {{1 \over {36}} + {1 \over {27}}}

= \sqrt {{{3 + 4} \over {9 \times 12}}}

= \sqrt {{7 \over {108}}} = {1 \over 6}\sqrt {{7 \over 3}}

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