A) \[1\]
B) \[\sqrt{\frac{a-x}{x-b}}\]
C) \[\sqrt{(a-x)(x-b)}\]
D) \[\frac{1}{\sqrt{(a-x)(b-x)}}\]
Correct Answer: B
Solution :
Let\[x=a{{\cos }^{2}}\theta +b{{\sin }^{2}}\theta \] \[\therefore \]\[a-x=a-a\cos \theta -b{{\sin }^{2}}\theta =(a-b){{\sin }^{2}}\theta \] and\[x-b=a{{\cos }^{2}}\theta -b{{\sin }^{2}}\theta -b=(a-b){{\cos }^{2}}\theta \] \[\therefore \]\[y=(a-b)\sin \theta \cdot cos\theta -(a-b){{\tan }^{-1}}\tan \theta \] \[=\frac{a-b}{2}\sin 2\theta -(a-b)\theta \] \[\therefore \]\[\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{(a-b)\cos 2\theta -(a-b)}{(b-a)\sin 2\theta }\] \[=\frac{1-\cos 2\theta }{\sin \theta }=\tan \theta =\sqrt{\frac{a-x}{x-b}}\]
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