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# Find the equation of rectangular hyperbola which has for one of its asymptotes the line $x+2y-5=0$ and passes through the point $(6 , 0 )$ and $(-3 , 0 )$ find its equation and asymptotes

Toolbox:
• (i) The asymptotes of $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1 are \large\frac{x}{a}$$\pm\large\frac{y}{b}$$=0. • (ii) They pass through the centre of the hyperbola. • (iii) Their slopes are \large\frac{b}{a} and -\large\frac{b}{a}( i.e)the axes of the hyperbola bisect the angles between them. • (iv) The angle between the asymptotes 2\alpha=2\tan^{-1}\large\frac{b}{a}$$=2\sec^{-1}e$
• In a rectangular hyperbola the asymptotes are at right angles a=b and $e=\sqrt 2$
Step 1:
One asymptotes of the rectangular hyperbola is $x+2y-5=0$.The other asymptote is $\perp$ to it.The equation is of the form $2x-y+l=0.$
The combined equation is $(x+2y-5)(2x-y+l)=0$
Step 2:
The equation of the hyperbola differs only in the constant term.
Therefore its equation is of the form $(x+2y-5)(2x-y+l)=k$------(1)
Step 3:
The rectangular hyperbola passes through $(6,0)$ and $(-3,0)$.
Substituting these in equ(1) we get
$(6-5)(12+l)=k$
$\Rightarrow k-l=12$----(2)
$(-3-5)(-6+l)=k$
$\Rightarrow k+8l=48$----(3)
Step 4:
Solving eq(2) & eq(3) we get
$9l=36$
$l=4$
Substitute the value of $l$ in eq(2) we get
$k-4=12$
$k=12+4$
$k=16$
Step 5:
The equation of the rectangular hyperbola is $(x+2y-8)(2x-y+4)=16$
On simplifying we get,
$2x^2+3xy-2y^2-6x+13y-36=0$
This is the required equation.