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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
If $\sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} = a(x - y)$, then ${{dy} \over {dx}} = $
  • A
    $\sqrt {{{1 - {x^2}} \over {1 - {y^2}}}} $
  • ✓
    $\sqrt {{{1 - {y^2}} \over {1 - {x^2}}}} $
  • C
    $\sqrt {{{{x^2} - 1} \over {1 - {y^2}}}} $
  • D
    $\sqrt {{{{y^2} - 1} \over {1 - {x^2}}}} $
✓

Answer

Correct option: B.
$\sqrt {{{1 - {y^2}} \over {1 - {x^2}}}} $
b
(b) Putting $x = \sin \theta $ and $y = \sin \phi $

$\cos \theta + \cos \phi= a(\sin \theta - \sin \phi)$

==> $2\cos \frac{{\theta + \phi}}{2}\cos \frac{{\theta - \phi}}{2} = a\left\{ {2\cos \frac{{\theta + \phi}}{2}\sin \frac{{\theta - \phi}}{2}} \right\}$

==> $\frac{{\theta - \phi}}{2} = {\cot ^{ - 1}}a \Rightarrow \theta - \phi = 2{\cot ^{ - 1}}a$

==> ${\sin ^{ - 1}}x - {\sin ^{ - 1}}y = 2{\cot ^{ - 1}}a$

==> $\frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{\sqrt {1 - {y^2}} }}\frac{{dy}}{{dx}} = 0 $

$\Rightarrow \frac{{dy}}{{dx}} = \sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}} $.

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