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Q)

Find the equation of the asymptotes of the following rectangular hyperbola $6x^{2}+5xy-6y^{2}+12x+5y+3=0$

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A)
Toolbox:
  • http://clay6.com/mpaimg/4_toolbox%2021.jpg
  • For the standard rectangular hyperbola,the coordinate axes are taken to the asymptotes and the equation of the hyperbola is $xy=c^2$ .
  • If the centre is at $(h,k)$ with asymptotes parallel to the coordinate axes,the equation is $(x-h)(y-k)=c^2$
Step 1:
$6x^2+5xy-6y^2+12x+5y+3=0$
Factorizing the 2nd degree terms,
$6x^2+5xy-6y^2=6x^2+9xy-4xy-6y^2$
$\qquad\qquad\qquad\;\;=3x(2x+3y)-2y(2x+3y)$
$\qquad\qquad\qquad\;\;=(3x-2y)(2x+3y)$
Step 2:
Therefore the equations of the asymptotes are of the form $3x-2y+l=0,2x+3y+m=0$
Then combined equation is $(3x-2y+l)(2x+3y+m)=0$----(1)
Step 3:
The combined equation of the asymptotes differs from that of the hyperbola only in the constant term.
Therefore their combined equation is of the form $6x^2+5xy-6y^2+12x+5y+3=0$------(2)
Step 4:
Eq(1) & Eq(2) represent the same pair of straight lines.
Comparing the coefficients of $x$ and $y$ we have,
$3m+2l=12$
$-2m+3l=5$
$\Delta=\begin{vmatrix}3 & 2\\-2 & 3\end{vmatrix}=9+4=13$
$\Delta_m=\begin{vmatrix}12 & 2\\5 & 3\end{vmatrix}=36-30=26$
$\Delta_l=\begin{vmatrix}3 & 12\\-2 & 5\end{vmatrix}=15+24=39$
Step 5:
$m=\large\frac{\Delta_m}{\Delta}=\large\frac{26}{13}$$=2$
$l=\large\frac{\Delta_l}{\Delta}=\large\frac{39}{13}$$=3$
The equations of the asymptotes are $3x-2y+3=0$,$2x+3y+2=0$
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