JEE MAIN - Mathematics (2019 - 10th January Morning Slot - No. 18)
The shortest distance between the point $$\left( {{3 \over 2},0} \right)$$ and the curve y = $$\sqrt x $$, (x > 0), is -
$${{\sqrt 3 } \over 2}$$
$${5 \over 4}$$
$${3 \over 2}$$
$${{\sqrt 5 } \over 2}$$
Explanation
Let points $$\left( {{3 \over 2},0} \right),\left( {{t^2},t} \right),t > 0$$
Distance = $$\sqrt {{t^2} + {{\left( {{t^2} - {3 \over 2}} \right)}^2}} $$
= $$\sqrt {{t^4} - 2{t^2} + {9 \over 4}} = \sqrt {{{\left( {{t^2} - 1} \right)}^2} + {5 \over 4}} $$
So minimum distance is $$\sqrt {{5 \over 4}} = {{\sqrt 5 } \over 2}$$
Distance = $$\sqrt {{t^2} + {{\left( {{t^2} - {3 \over 2}} \right)}^2}} $$
= $$\sqrt {{t^4} - 2{t^2} + {9 \over 4}} = \sqrt {{{\left( {{t^2} - 1} \right)}^2} + {5 \over 4}} $$
So minimum distance is $$\sqrt {{5 \over 4}} = {{\sqrt 5 } \over 2}$$
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