A) \[\frac{\sqrt{5}}{2}\]
B) \[\sqrt{5}\]
C) \[2\sqrt{5}\]
D) \[\sqrt{2}\]
Correct Answer: B
Solution :
Given curve is \[{{y}^{2}}(2-x)={{x}^{3}}\Rightarrow {{y}^{2}}=\frac{{{x}^{3}}}{2-x}\] On differentiating w. r. t. x, we get \[2y\frac{dy}{dx}=(2-x)\frac{d}{dx}({{x}^{3}})-\frac{{{x}^{3}}\frac{d}{dx}(2-x)}{{{(2-x)}^{2}}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{2y}\left[ \frac{(2-x)(3{{x}^{2}})-{{x}^{3}}(-1)}{{{(2-x)}^{2}}} \right]\] \[=\frac{1}{2y}\left[ \frac{6{{x}^{2}}-3{{x}^{2}}+{{x}^{3}}}{{{(2-x)}^{2}}} \right]=\frac{1}{2y}\left[ \frac{6{{x}^{2}}-2{{x}^{3}}}{{{(2-x)}^{2}}} \right]\] At point \[(1,1),\frac{dy}{dx}=\frac{1}{2\times 1}\left[ \frac{6\times {{(1)}^{2}}-2\times {{(1)}^{3}}}{{{(2-1)}^{2}}} \right]\] \[=\frac{1}{2}\left[ \frac{6-2}{1} \right]=\frac{4}{2}=2\] So, the length of normal \[=y\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}\] \[=1\sqrt{1+{{(2)}^{2}}}=1\sqrt{1+4}=\sqrt{5}\]
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