The locus of points(x,y) in the plane satisfying sin2x + sin2y =1 consists of
- a circle centered at origin
- infinitely many circles that are all centered at the origin
- infinitely many lines with slope ±1
- finitely many lines with slope ±1
The Correct Option is C
Approach Solution - 1
The given equation is x2 + 2xsin(xy) + 1 = 0.
Solving for sin(xy), we find that it is equal to \(\frac{-2x}{(1+x^2)}\), which can also be expressed as \(\frac{-2}{(x+1)}\).
Since (x+1) is always greater than or equal to 2, it follows that sin(xy) is less than or equal to -1. This situation is possible only when sin(θ) equals -1 for some angle θ.
Therefore, we have sin(xy) = -1, which implies that xy =\( \frac{-π}{2}\).
This equation describes a hyperbola because it is in the form of xy = a constant (in this case, a negative constant, \(\frac{-π}{2}\)).
So, the correct answer is option (C): infinitely many lines with slope ±1.
Approach Solution -2
Given :
\(\sin^2x+\sin^2y=1\)
\(⇒\sin^2y=1-\sin^2x\)
\(⇒\sin^2y=\cos^2x\)
So, \(\sin y=±\cos x\)
Now, if \(\sin y=\cos x=\sin(\frac{\pi}{2}-x)\)
Then, \(y=n\pi+(-1)^n(\frac{\pi}{2}-x),n\in Z\)
which represents equation of line with slope ±1
And Now, if
\(\sin y=-\cos x=-\sin(\frac{\pi}{2}-x)=\sin(x-\frac{\pi}{2})\)
\(⇒y=n\pi+(-1)^n(x-\frac{\pi}{2}),n\in Z\)
which represents equation of line with slope ±1
So, the correct option is (C) : infinitely many lines with slope ±1.
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Concepts Used:
Angle between Two Planes
A plane is demarcated as two-dimensional in nature and the one which has a flat surface that prolongs infinitely far in two dimensions. It is set up by some stack of lines that are kept together.
Angles between two planes refer to the acute angle which is manifest by the standard vectors of the planes. If the standard vectors of two planes are rectangular, then we can say they are perpendicular. This specific portion tells us what a plane is and what is the angle between the two planes, it also shows us how to calculate the angles between the two planes in a Cartesian plane.