A) \[x=\frac{1}{4}\]
B) \[y=\frac{1}{4}\]
C) \[y=\frac{1}{2}\]
D) \[y=1\]
Correct Answer: C
Solution :
Let the equation of line parallel to x-axis be \[y=\lambda \] .....(i) Solving (i) with the cuve \[y=\sqrt{x}\] .....(ii) We get \[P({{\lambda }^{2}},\,\,\lambda )\] the point of intersection at P \[\therefore \] Slope of (ii) is, m= \[{{\left( \frac{dy}{dx} \right)}_{\text{at }P}}=\frac{1}{2\lambda }\] \[\therefore \] (i) and (ii) intersect at P, at \[45{}^\circ \] \[\therefore \] \[{{\tan }^{-1}}\,\left( \frac{m-0}{1+m.0} \right)=\pm 45{}^\circ \]. Þ \[m=\left( \frac{1}{2\lambda } \right)=\pm \,1\] Þ \[\lambda =\pm \,\frac{1}{2}\] \[\therefore \] The equation of line is \[y=\frac{1}{2}\] or \[y=\frac{-1}{2}\] but \[y=\frac{-1}{2}\] is not given, hence the required line is \[y=\frac{1}{2}\].
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