A) \[xy'-\,\sqrt{1-{{x}^{2}}}=\,0\]
B) \[y\,=\,-\,\text{lo}{{\text{g}}_{e}}\,\left( \frac{1+\sqrt{1-{{x}^{2}}}}{x} \right)+\,\sqrt{1-{{x}^{2}}}\]
C) \[xy'+\sqrt{1-{{x}^{2}}}=\,0\]
D) \[y\,=\,\text{lo}{{\text{g}}_{e}}\,\left( \frac{1+\sqrt{1\,\,-\,{{x}^{2}}}}{x} \right)-\,\sqrt{1-{{x}^{2}}}\]
Correct Answer: C , D
Solution :
Equation of Tangent at P |
\[y-y=\frac{dy}{dx}\,(X-x)\] |
For \[{{\text{Y}}_{p}}\]\[\Rightarrow \]\[(X=0)\] |
\[{{Y}_{p}}=y-x\frac{dy}{dx}\] |
Distance \[{{\text{Y}}_{\text{p}}}\text{P=1}\] |
\[{{x}^{2}}+{{\left( y-y+x\frac{dy}{dx} \right)}^{2}}=1\] |
\[{{x}^{2}}\left( 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right)\,=\,1\] |
\[{{\left( \frac{dy}{dx} \right)}^{2}}=\frac{1}{{{x}^{2}}}-1\] |
\[\frac{dy}{dx}=\pm \frac{\sqrt{1-x}}{x}\to \]option A and C |
\[\int{dy}=\pm \int{\frac{\sqrt{1-{{x}^{2}}}}{x}dx}\] |
\[x=\text{sin}\theta \] |
\[y=\pm \int{\frac{\cos \theta }{\sin \theta }}\cos \theta d\theta \] |
\[y=\pm \int{\frac{1-{{\sin }^{2}}\theta }{\sin \theta }}d\theta \] |
\[y=\pm \int{(cosec\theta -sin\theta )d\theta }\] |
\[y=\pm (In\left| \cos ec\theta +\cot \theta \left| +\cos \theta \right. \right.)+\text{C}\] |
P as (1, 0) \[\Rightarrow \]\[C=0\] |
\[y=\pm \left( \text{In}\left( \frac{1+\sqrt{1-{{x}^{2}}}}{x} \right) \right)+\sqrt{1-{{x}^{2}}}\] |
\[\to \]option [b], [d], |
Solution :
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