Single Choice

The shortest distance between the line $$y-x =1$$ and the curve $$\mathrm{x}=\mathrm{y}^{2}$$ is

A$$\displaystyle \frac{3\sqrt{2}}{8}$$
Correct Answer
B$$\displaystyle \frac{2\sqrt{3}}{8}$$
C$$\displaystyle \frac{3\sqrt{2}}{5}$$
D$$\displaystyle \frac{\sqrt{3}}{4}$$

Solution

The shortest distance between the curves will between the two points where the tangents are parallel to each other.
Hence,
$$1=2\displaystyle \mathrm{y}\dfrac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$$

$$\Rightarrow \displaystyle \dfrac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\dfrac{1}{2\mathrm{y}}=1$$

$$= \displaystyle \mathrm{y}=\dfrac{1}{2}= \displaystyle \mathrm{x}=\dfrac{1}{4}$$

The distance of $$\left(\dfrac12, \dfrac14 \right) $$ from the line $$y-x =1 $$ will be the shortest distance.
Shortest distance $$=|\displaystyle \dfrac{\dfrac{1}{2}-\dfrac{1}{4}-1}{\sqrt{2}}|=\dfrac{3}{4\sqrt{2}}=\dfrac{3\sqrt{2}}{8}$$
Hence, option 'A' is correct.

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