All the points (x, y) in the plane satisfying | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation: $x^2+2x \sin(xy)+1=0$We can rewrite this as: $x^2+2x \sin(xy)+\sin^2(xy)+1-\sin^2(xy)=0$This simplifies to: $(x+\sin(xy))^2+\c

Vedclass · Accepted Answer

a pair of straight lines

Vedclass · Answer

(A) Given the equation: $x^2+2x \sin(xy)+1=0$We can rewrite this as: $x^2+2x \sin(xy)+\sin^2(xy)+1-\sin^2(xy)=0$This simplifies to: $(x+\sin(xy))^2+\cos^2(xy)=0$Since the sum of two squares is zero,each term must be zero:$(x+\sin(xy))^2=0 \Rightarrow x+\sin(xy)=0$$\cos^2(xy)=0 \Rightarrow \cos(xy)=0$From $\cos(xy)=0$,we have $xy = (2n+1)\frac{\pi}{2}$ for some integer $n$.Substituting $\sin(xy) = \pm 1$ into $x+\sin(xy)=0$,we get $x = \mp 1$.If $\sin(xy)=1$,then $x=-1$. Since $xy = (2n+1)\frac{\pi}{2}$,we get $y = -(2n+1)\frac{\pi}{2}$.If $\sin(xy)=-1$,then $x=1$. Since $xy = (2n+1)\frac{\pi}{2}$,we get $y = (2n+1)\frac{\pi}{2}$.These represent a set of parallel lines $x=1$ and $x=-1$,which are a pair of straight lines.

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

Similar Questions

If two of the lines represented by $2x^3+x^2y+y^3=0$ are mutually perpendicular,then the slope of the third line is

The joint equation of the pair of lines passing through $(2,3)$ and parallel to the lines represented by $x^{2}-y^{2}=0$ is:

The product of the perpendicular distances drawn from the origin to the pair of straight lines $6x^2 - 5xy - 6y^2 + x + 5y - 1 = 0$ is

If $s$ and $p$ are respectively the sum and the product of the slopes of the lines $3x^2 - 2xy - 15y^2 = 0$,then $s:p$ is equal to

The area of the triangle formed by the lines $x^2 - 4y^2 = 0$ and $x = a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module