If a-2bcos x a-2b cos y -a2-b2- where a-b-0- then dxdy at -4-4 is

( a+√(2)bcos x ) ( a √(2)b cos y )=a^2 b^2 Differentiation w.r.t y, we get ( √(2)bsin x·dxdy) ( a √(2)b cos y nbsp; )+ ( a+√(2)b cos x ) ( √(2)b sin y n ...

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Byju's Answer

Standard XII

Mathematics

Variable Separable Method

If a+√2bcos x...

Question

If $(a + \sqrt{2} b cos x) (a - \sqrt{2} b cos y) = a^{2} - b^{2}$ , where $a > b > 0$ , then $\frac{d x}{d y}$ at $(\frac{π}{4}, \frac{π}{4})$ is

\frac{a + b}{a - b}

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\frac{a - 2 b}{a + 2 b}

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\frac{a - b}{a + b}

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\frac{2 a + b}{2 a - b}

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Solution

The correct option is A $\frac{a + b}{a - b}$
$(a + \sqrt{2} b cos x) (a - \sqrt{2} b cos y) = a^{2} - b^{2}$
Differentiation w.r.t $y,$ we get
$(- \sqrt{2} b sin x \cdot \frac{d x}{d y}) (a - \sqrt{2} b cos y) + (a + \sqrt{2} b cos x) (\sqrt{2} b sin y) = 0$
At $x = y = \frac{π}{4},$ we have
$\Rightarrow - \frac{d x}{d y} (a - b) b + (a + b) (b) = 0$
$∴ \frac{d x}{d y} = \frac{a + b}{a - b}$

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If (a+√2bcosx)(a−√2bcosy)=a2−b2, where a>b>0, then dxdy at (π4,π4) is

The correct option is A a+ba−b(a+√2bcosx)(a−√2bcosy)=a2−b2 Differentiation w.r.t y, we get (−√2bsinx⋅dxdy)(a−√2bcosy)+(a+√2bcosx)(√2bsiny)=0 At x=y=π4, we have ⇒−dxdy(a−b)b+(a+b)(b)=0 ∴dxdy=a+ba−b

If $(a + \sqrt{2} b cos x) (a - \sqrt{2} b cos y) = a^{2} - b^{2}$ , where $a > b > 0$ , then $\frac{d x}{d y}$ at $(\frac{π}{4}, \frac{π}{4})$ is