Question & Answer
QUESTION

Find the asymptotes of xy-3y-2x=0.

ANSWER Verified Verified
Hint: In analytic geometry, an asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity. In projective geometry and related contexts, an asymptote of a curve is a line which is tangent to the curve at a point at infinity.

Complete Step-by-Step solution:
There are three kinds of asymptotes: horizontal, vertical and oblique. For curves given by the graph of a function y = ƒ(x), horizontal asymptotes are horizontal lines that the graph of the function approaches as x tends to \[+\infty \ to\ -\infty \]. Vertical asymptotes are vertical lines near which the function grows without bound. An oblique asymptote has a slope that is non-zero but finite, such that the graph of the function approaches it as x tends to \[+\infty \ to\ -\infty \] .

As it is mentioned in this question, we have to find the asymptotes of the given equation which is of a hyperbola.
Now, as the equation of a hyperbola and its asymptotes are different from each other because of a constant term only. Therefore, the equation of pair of asymptotes is given as follows
 xy – 3y – 2x + λ = 0 (Where λ is any constant such that it represents two straight lines)
Now, comparing this equation with general equation of a conic that is as follows
\[abc+2fgh-a{{f}^{2}}-\text{ }b{{g}^{2}}-c{{h}^{2}}=\text{ }0~\]
Therefore, the value of \[\lambda \] can be found out as \[\lambda =6\ \ \ \ \ ...(1)\] .
 From (1), we get that the asymptotes of given hyperbola are given as follows
\[\begin{align}
  & xy-3y-2x+6\text{ }=\text{ }0 \\
 & \left( y-2 \right)\cdot \left( x-3 \right)=\text{ }0~ \\
\end{align}\]
 Hence, the equation of asymptotes are x – 3 = 0 and y – 2 = 0.

Note: The students can make an error if they don’t know about the general formula of a conic and if they are not able to find out that the given equation is of a hyperbola. These two things are very important to solve this question.