A) It makes a constant angle with x-axis
B) It passes through the origin
C) It is at a constant distance from the origin
D) None of these
Correct Answer: C
Solution :
\[y=a(\sin \theta -\theta \cos \theta ),\,x=a(\cos \theta +\theta \sin \theta )\] \[\frac{dy}{d\theta }=a[\cos \theta -\cos \theta +\theta \sin \theta ]=a\theta \sin \theta \] \[\frac{dx}{d\theta }=a(-\sin \theta +\sin \theta +\theta \cos \theta )=a\theta \cos \theta \] \[\therefore \,\,\,\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{a\theta \sin \theta }{a\theta \cos \theta }=\tan \theta \] Þ Slope of the tangent = \[\tan \theta \] \ Slope of the normal = \[-\cot \theta \] Hence, equation of normal \[[y-a\sin \theta +a\theta \cos \theta ]=-\frac{\cos \theta }{\sin \theta }\]\[[x-a\cos \theta -a\theta \sin \theta ]\] Þ \[y\sin \theta -a{{\sin }^{2}}\theta +a\theta \sin \theta \cos \theta \] \[=-x\cos \theta +a{{\cos }^{2}}\theta +a\theta \sin \theta \cos \theta \] Þ \[x\cos \theta +y\sin \theta =a({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )\] Þ \[x\cos \theta +y\sin \theta =a\] \Distance from origin = \[\frac{a}{\sqrt{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }}=\text{a constant}\]
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