Question

The locus of points(x,y) in the plane satisfying sin²x + sin²y =1 consists of

• A. a circle centered at origin
• B. infinitely many circles that are all centered at the origin
• C. infinitely many lines with slope ±1
• D. finitely many lines with slope ±1

Answer 1

The Correct Option is C

Given equation: $x^2 + 2x \sin(xy) + 1 = 0$

Rewriting, we get: 

$\Rightarrow 2x \sin(xy) = - (x^2 + 1)$

$\Rightarrow \sin(xy) = \frac{ - (x^2 + 1)}{2x}$

Now simplify:

$\sin(xy) = \frac{ -2x }{x^2 + 1}$

Now, observe that for $\sin(xy) = -1$, we must have:

$\frac{ -2x }{x^2 + 1} = -1$

$\Rightarrow \frac{2x}{x^2 + 1} = 1$

$\Rightarrow 2x = x^2 + 1$

$\Rightarrow x^2 - 2x + 1 = 0$

$\Rightarrow (x - 1)^2 = 0 \Rightarrow x = 1$

Now substitute back to find $y$:

$\sin(xy) = -1 \Rightarrow xy = -\frac{\pi}{2} \Rightarrow y = -\frac{\pi}{2}$ (since $x = 1$)

But, the equation $\sin(xy) = -1$ has infinitely many solutions:

$xy = -\frac{\pi}{2}, -\frac{3\pi}{2}, -\frac{5\pi}{2}, \ldots$

Each such equation defines a branch of the hyperbola $xy = \text{constant}$

Each branch is asymptotic to lines of slope ±1.

Hence, the correct answer is: (C) infinitely many lines with slope ±1.

Answer 2

Given: $\sin^2 x + \sin^2 y = 1$ 

$\Rightarrow \sin^2 y = 1 - \sin^2 x$

$\Rightarrow \sin^2 y = \cos^2 x$

$\Rightarrow \sin y = \pm \cos x$

Case 1: $\sin y = \cos x = \sin\left(\frac{\pi}{2} - x\right)$

Then: $y = n\pi + (-1)^n\left(\frac{\pi}{2} - x\right), \; n \in \mathbb{Z}$

This represents a family of lines with slope ±1

Case 2: $\sin y = -\cos x = -\sin\left(\frac{\pi}{2} - x\right) = \sin\left(x - \frac{\pi}{2}\right)$

Then: $y = n\pi + (-1)^n\left(x - \frac{\pi}{2}\right), \; n \in \mathbb{Z}$

This also represents a family of lines with slope ±1

Hence, the correct answer is: (C): infinitely many lines with slope ±1