JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 20)
A helicopter is flying along the curve given by y – x3/2 = 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 -
$${1 \over 6}\sqrt {{7 \over 3}} $$
$${{\sqrt 5 } \over 6}$$
$${1 \over 2}$$
$${1 \over 3}$$$$\sqrt {{7 \over 3}} $$
Explanation
_10th_January_Evening_Slot_en_20_1.png)
$$y - {x^{3/2}} = 7\left( {x \ge 0} \right)$$
$${{dy} \over {dx}} = {3 \over 2}{x^{1/2}}$$
$$\left( {{3 \over 2}\sqrt x } \right)\left( {{{7 - y} \over {{1 \over 2} - x}}} \right) = - 1$$
$$\left( {{3 \over 2}\sqrt x } \right)\left( {{{ - {x^{3/2}}} \over {{1 \over 2} - x}}} \right) = - 1$$
$${3 \over 2}.{x^2} = {1 \over 2} - x$$
$$3{x^2} = 1 - 2x$$
$$3{x^2} + 2x - 1 = 0$$
$$3{x^2} + 3x - x - 1 = 0$$
$$\left( {x + 1} \right)\left( {3x - 1} \right) = 0$$
$$ \therefore $$ $$x = - 1$$ (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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