JEE PYQ: Application Of Derivatives Question 19
Question 19 - 2021 (25 Feb Shift 2)
The shortest distance between the line $x - y = 1$ and the curve $x^2 = 2y$ is:
(1) $\frac{1}{2}$
(2) $0$
(3) $\frac{1}{2\sqrt{2}}$
(4) $\frac{1}{\sqrt{2}}$
Show Answer
Answer: (3)
Solution
The tangent to $x^2 = 2y$ with slope 1 is $y = x - \frac{1}{2}$, i.e., $x - y = \frac{1}{2}$. Distance from $x - y = 1$: $\frac{|1 - 1/2|}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.
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