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If $ \sqrt{1-y^2}-\sqrt{1-x^2}=a(x-y),\normalsize$ then find $\large\frac{dy}{dx}$

$\begin{array}{1 1}(a)\;\sqrt{\large\frac{1-x^2}{1-y^2}}\\(b)\; -\sqrt{\large\frac{1-x^2}{1-y^2}}\\(c)\;\sqrt{\large\frac{1-y^2}{1-x^2}}\\(d)\;-\sqrt{\large\frac{1-y^2}{1-x^2}}\end{array}$

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1 Answer

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Toolbox:
  • $\sin x-\sin y=2\cos(\large\frac{x+y}{2})$$\sin(\large\frac{x-y}{2})$
  • $\cos x-\cos y=-2\sin(\large\frac{x+y}{2})$$\sin(\large\frac{x-y}{2})$
Step 1
$ \sqrt {1-y^2}-\sqrt{1-x^2}=a(x-y)$
Put $x=\sin\alpha\:\;and\:\:y=\sin\beta$
$\Rightarrow\:\alpha=\sin^{-1}x\:\:and\:\:\beta=\sin^{-1}y$
$\Rightarrow\: \sqrt {1-\sin^2\alpha}-\sqrt{1-\sin^2\beta}=a(\sin\alpha-\sin\beta)$
$\Rightarrow\:\cos \alpha- \cos\beta=a(\sin\alpha-\sin\beta)$
Step 2
$\Rightarrow\:-2\sin(\large\frac{\alpha+\beta}{2})$$\sin(\large\frac{\alpha-\beta}{2})$$=2a\:\cos(\large\frac{\alpha+\beta}{2})$$\sin(\large\frac{\alpha-\beta}{2})$
$\Rightarrow\:tan(\large\frac{\alpha+\beta}{2})$$=-a$
$\Rightarrow\:\alpha+\beta=\tan^{-1}(-2a)$
Step 3
Substituting the value of $\alpha\:\:and\:\:\beta$,
$\Rightarrow\:sin^{-1}x+\sin^{-1}y=\tan^{-1}(-2a)$
Step 4
Differentiating both the sides w.r.to $x$
$\large\frac{1}{\sqrt{1-x^2}}$$+\large\frac{1}{\sqrt{1-y^2}}.\frac{dy}{dx}$$=0$
$\Rightarrow\:\large\frac{dy}{dx}=-\sqrt{\large\frac{1-y^2}{1-x^2}}$

 

answered Feb 5, 2014 by rvidyagovindarajan_1
edited Mar 24, 2014 by meena.p
In step 2, instead of  α + β = arc(tan (-2a))
 α + β = 2 arc(tan(-a)) must be there
 

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