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Q)

Find the equation of the hyperbola if its asymptotes are parallel to $x+2y-12=0$ and $x-2y+8=0, (2 , 4 ) $ is the centre of the hyperbola and it passes through $(2 , 0 )$

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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 asymptotes are parallel to $x+2y-12=0$ (slope=$\large\frac{-1}{2}$) and $x-2y+8=0$ (slope=$\large\frac{1}{2})$
They pass through the centre $(2,4)$ of the hyperbola.
Step 2:
Therefore their equations are of the form $(y-y_1)=m(x-x_1)$
Where $(x_2,y_1)=(2,4)$
For $m=\large\frac{-1}{2}$
$(y-4)=-\large\frac{-1}{2}$$(x-2)$
$\Rightarrow 2y-8=-x+2$
$\Rightarrow x+2y-10=0$
For $m=\large\frac{1}{2}$
$(y-4)=\large\frac{1}{2}$$(x-2)$
$\Rightarrow 2y-8=x-2$
$\Rightarrow x-2y+6=0$
The combined equation of the asymptotes is $(x+2y-10)(x-2y+6)=0$
Step 3:
The equation of the hyperbola differs only in the constant term.Let it be $(x+2y-10)(x-2y+6)=k$
The hyperbola passes through $(2,0)$
Therefore $(2-10)(2+6)=k$
$\Rightarrow k=-64$
The equation of the hyperbola is $(x+2y-10)(x-2y+6)=-64$
On simplifying we get,
$x^2-4y^2-4x+32y+4=0$
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