Question

The equations of the lines represented by the equation $ab\left( {{x}^{2}}-{{y}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right)xy=0$ are
A. $ax-by=0,bx+ay=0$
B. $ax-by=0,bx+ay=0$
C. $ax+by=0,bx+ay=0$
D. $ax+by=0,bx-ay=0$

Answer 1

Hint:The equation $ab\left( {{x}^{2}}-{{y}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right)xy=0$ represents the pair of straight lines. It contains two separate lines. By simplifying the equation $ab\left( {{x}^{2}}-{{y}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right)xy=0$ we can easily find the equations of the separate lines.

Formula Used$a{{x}^{2}}+2hxy+b{{y}^{2}}$




Complete step by step solution:The given equation of the pair of straight line is given as $ab\left( {{x}^{2}}-{{y}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right)xy=0$
To determine the equation of the separate lines we have to simplify the given equation of the pair of straight lines as follows-
The given equation is $ab\left( {{x}^{2}}-{{y}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right)xy=0$.
Multiplying $ab$ with $\left( {{x}^{2}}-{{y}^{2}} \right)$ and $xy$ with $\left( {{a}^{2}}-{{b}^{2}} \right)$ we get-
$
   ab\left( {{x}^{2}}-{{y}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right)xy=0 \\
  ab{{x}^{2}}-ab{{y}^{2}}+{{a}^{2}}xy-{{b}^{2}}xy=0 \\ $
Taking $ax,by$ as a common factor we get the result as-
$
   ab{{x}^{2}}-ab{{y}^{2}}+{{a}^{2}}xy-{{b}^{2}}xy=0 \\
  ax\left( bx+ay \right)-by\left( ay+bx \right)=0 \\
  \left( bx+ay \right)\left( ax-by \right)=0 \\
$
Thus the equations of the separate lines are given as $
   \left( bx+ay \right)=0 \\
  \left( ax-by \right)=0 \\
$ .
Thus we can write that the equations of the lines represented by the equation $ab\left( {{x}^{2}}-{{y}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right)xy=0$ are $ax-by=0,bx+ay=0$.


Option ‘A’ is correct

Additional Information:The general form of the equation of a pair of straight lines is given as $a{{x}^{2}}+2hxy+b{{y}^{2}}$_.
Comparing it with the given equation $ab\left( {{x}^{2}}-{{y}^{2}} \right)+\left( {{a}^{2}}-{{b}^{2}} \right)xy=0$ we get the values of the coefficients as a as ab, b as –ab and $h$ as $\left( {{a}^{2}}-{{b}^{2}} \right)/2$ .





Note: The another method to solve this equation is to determine the sum of the coefficients $a,b$ . If the sum of $a,b$ is zero then the straight lines are perpendicular to each other.