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  • Minimum distance between the curves <img alt="\mathrm{y}^{2}=\mathrm{x}-1$ and $\mathrm{x}^{2}=\mathrm{y}-1" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7By%7D%5E%7B2%7D%3D%5Cmat

Minimum distance between the curves \mathrm{y}^{2}=\mathrm{x}-1$ and $\mathrm{x}^{2}=\mathrm{y}-1  is equal to

Option: 1

\frac{3 \sqrt{2}}{4}


Option: 2

\frac{5 \sqrt{2}}{4}


Option: 3

\frac{7}{4} \sqrt{2}


Option: 4

\frac{1}{4} \sqrt{2}


Answers (1)

best_answer

Both curves are symmetrical about the line \mathrm{y=x}. If line \mathrm{AB} is the line of shortest distance then at A and B slopes of curves should be equal to one.

\mathrm{For\: \mathrm{y}^{2}=\mathrm{x}-1, \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}}=1}
\mathrm{\Rightarrow \quad y=\frac{1}{2}, x=\frac{5}{4}}
\mathrm{\Rightarrow \quad \mathrm{B} \equiv\left(\frac{1}{2}, \frac{5}{4}\right)}.


Hence minimum distance \mathrm{AB},
\mathrm{=\sqrt{\left(\frac{5}{4}-\frac{1}{2}\right)^{2}+\left(\frac{5}{4}-\frac{1}{2}\right)^{2}}=\frac{3 \sqrt{2}}{4}} units.
 

Posted by

Ritika Jonwal

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