A) \[\frac{\sqrt{5}}{6}\]
B) \[\frac{1}{3}\sqrt{\frac{7}{3}}\]
C) \[\frac{1}{6}\sqrt{\frac{7}{3}}\]
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
| \[y={{x}^{3/2}}-2\] \[\frac{dy}{dx}=\frac{3}{2}\sqrt{x}\] |
| Slope of normal \[=-\frac{2}{3\sqrt{x}}\] |
| Let point is\[({{x}_{1}},x_{1}^{3/2}-2)\] |
| \[\therefore \]Normal\[y-(x_{1}^{3/2}-2)=\frac{-2}{3\sqrt{{{x}_{1}}}}(x-{{x}_{1}})\] |
| Now put\[(1,7)\]and solve it. \[\Rightarrow \] \[{{x}_{1}}=\frac{1}{3}\] |
| \[\therefore \] \[P\Rightarrow \left( \frac{1}{3},7+\frac{1}{3\sqrt{3}} \right),A\Rightarrow (1,7)\] |
| \[\therefore \] \[AD=\frac{1}{6}\sqrt{\frac{7}{3}}.\] |
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