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Q)

Find the equation of the asmptotes to the hyperbola $ 8x^{2}+10xy-3y^{2}-2x+4y-2=0$

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A)
Toolbox:
  • The equation of a hyperbola and the combined equation of the asymptotes differ only in the constant term.
Step 1:
The equation of a hyperbola and the combined equation of the asymptotes differ only in the constant term.
Therefore the combined equation of the asymptotes is of the form $8x^2+10xy-3y^2-2x+4y+k=0$-----(1)
Step 2:
Factorising the 2nd degree terms,
$8x^2+10xy-3y^2=8x^2+12xy-2xy-3y^2$
$\qquad\qquad\;\;\;\;\;\;\;\;\;\;\;=4x(2x+3y)-y(2x+3y)$
$\qquad\qquad\;\;\;\;\;\;\;\;\;\;\;=(4x-y)(2x+3y)$
Therefore the asymptotes have equations $4x-y+l=0,2x+3y+m=0$ and their combined equation is $(4x-y+l)(2x+3y+m)=0$------(2)
Step 3:
Equations (1) & (2) represent the same pair of lines.
Comparing the coefficient of $x$ and $y$
$4m+2l=-2$
$-m+3l=4$
$\Delta=\begin{vmatrix}4 & 2\\-1 & 3\end{vmatrix}=12+2=14$
$\Delta m=\begin{vmatrix}-2 & 2\\4 & 3\end{vmatrix}=-6-18=-14$
$\Delta l=\begin{vmatrix}4& -2\\-1 & 4\end{vmatrix}=16-2=14$
$l=\large\frac{\Delta l}{\Delta}$$=1$
$m=\large\frac{\Delta m}{\Delta}$$=-1$
Therefore the asymptotes are $2x+3y-1=0$ and $4x-y+1=0$ and their combined equation is $8x^2+16xy-3y^2-2x+4y-1=0$
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