If a+√2bcosx)(a-√2bcosy=a^2-b^2, where a>b>0, thendx/dy at (π)/4,(π)/4 is:

Correct answer: (C) (a+b)/(a-b). a+√2bcosxa−√2bcosy=a^2−b^2 a−√2bcosy−√2bsinxdx/dy+√2bsinya+√2bcosx=0 dx/dy=(√2bsinya+√2bcosx)/(√2bsinxa−√2bcosy)=(sinya+√2bcosx)/(sinxa−√2bcosy) dx/dy_(π4,π4)=(a+b)/(a−b).

Q34 / 75 · JEE Main 2020 (04 Sep Shift 1) · Mathematics

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:

A
$\frac{a - 2 b}{a + 2 b}$
B
$\frac{a - b}{a + b}$
Correct answer
C
$\frac{a + b}{a - b}$
D
$\frac{2 a + b}{2 a - b}$

Solution

$(a + \sqrt{2} b \cos x) (a - \sqrt{2} b \cos y) = a^{2} - b^{2}$

$(a - \sqrt{2} b \cos y) (- \sqrt{2} b \sin x) \frac{d x}{d y} + \sqrt{2} b \sin y (a + \sqrt{2} b \cos x) = 0$

$\frac{d x}{d y} = \frac{\sqrt{2} b \sin y (a + \sqrt{2} b \cos x)}{\sqrt{2} b \sin x (a - \sqrt{2} b \cos y)} = \frac{\sin y (a + \sqrt{2} b \cos x)}{\sin x (a - \sqrt{2} b \cos y)}$

$\frac{d x}{d y} |_{(\frac{π}{4}, \frac{π}{4})} = \frac{a + b}{a - b}$ .

Hence, the correct answer is option: 'C'.

If a+2bcosx)(a-2bcosy=a2-b2, where a>b>0, thendxdy at π4,π4 is:

Solution

Relevant Questions

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: