Question

All the points $(x,y)$ in the plane satisfying the equation $x^2+2x\sin \left(xy\right)+1=0$ lie on.

• A. A pair of straight lines
• B. A family of hyperbolas
• C. A parabola
• D. An ellipse

Answer 1

The correct option is B A family of hyperbolas

$x^2+2xsin\left(xy\right)+1=0$

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

$\because (\frac{1}{x}+x)⩾2$

$\Rightarrow sin\left(xy\right)⩽-1$

Since, this is possible only when $sin\left(\theta \right)=1$ for some $\theta$.

$\therefore sin\left(xy\right)=-1\Rightarrow xy=-\Pi /2$

This equation is of the form xy = negative constant, which means it is a hyperbola in 2nd and 4th quadrant.