JEE PYQ: Application Of Derivatives Question 75
Question 75 - 2019 (10 Jan Shift 2)
A helicopter is flying along the curve given by $y - x^{3/2} = 7$, $(x \ge 0)$. A soldier positioned at the point $\left(\frac{1}{2}, 7\right)$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is:
(1) $\frac{\sqrt{5}}{6}$
(2) $\frac{1}{3}\sqrt{\frac{7}{3}}$
(3) $\frac{1}{6}\sqrt{\frac{7}{3}}$
(4) $\frac{1}{2}$
Show Answer
Answer: (3)
Solution
Point on curve: $(x_1, x_1^{3/2} + 7)$. Slope of normal $= -1/m_{tangent}$. Normal must pass through $(\frac{1}{2}, 7)$. Solving: $x_1 = \frac{1}{3}$. Distance $= \sqrt{(\frac{1}{3}-\frac{1}{2})^2 + (\frac{1}{3\sqrt{3}})^2} = \sqrt{\frac{1}{36} + \frac{1}{27}} = \frac{1}{6}\sqrt{\frac{7}{3}}$.
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