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# Find the equations of the rectangular hyperbola which has for one its asymptotes the line $x+2y-5=0$ and passes through the points $\left( 6,0 \right)$ and $\left( -3,0 \right)$ 

Last updated date: 20th Jun 2024
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Hint:
We use the fact that if the equations of asymptotes of a hyperbola are given by ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ then the equation of the hyperbola is given by $\left( {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}} \right)\left( {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}} \right)=k$ . We are already given ${{c}_{1}}$ in this question we find ${{c}_{2}},k$ by using perpendicular condition on asymptotes of rectangular hyperbola and satisfaction of given points $\left( 6,0 \right)$ and $\left( -3,0 \right)$ .

We are given the equation one asymptote as
$x+2y-5=0$
We know that the asymptotes of rectangular hyperbolas are always perpendicular. We know that the equation of line perpendicular to $ax+by+c=0$ is given by $bx-ay+{{c}_{2}}=0$ where $c,{{c}_{2}}$ are constant terms. So the equation of the other asymptote is the equation of line perpendicular to given asymptote $x+2y-5=0$ which is
\begin{align} & 2\cdot x-1\cdot y+{{c}_{2}}=0 \\ & \Rightarrow 2x-y+{{c}_{2}}=0 \\ \end{align}
So we multiply respective sides of the equations of asymptotes and find the equation of rectangular hyperbola as
$\left( x+2y-5 \right)\left( 2x-y+{{c}_{2}} \right)=k.........\left( 1 \right)$
Here $k$ is a real constant. We are also given in the question that the rectangular hyperbola passes through the points $\left( 6,0 \right)$ and $\left( -3,0 \right)$ . So these points will satisfy the equation of rectangular hyperbola. We put $\left( 6,0 \right)$ in equation of hyperbola and have;
\begin{align} & \left( 6+2\cdot 0-5 \right)\left( 2\cdot 6-0+{{c}_{2}} \right)=k \\ & \Rightarrow 1\left( 12+{{c}_{2}} \right)=k \\ & \Rightarrow 12+{{c}_{2}}=k.......\left( 2 \right) \\ \end{align}
We put $\left( -3,0 \right)$ in equation of rectangular hyperbola (1) and have;
\begin{align} & \left( -3+2\cdot 0-5 \right)\left( 2\left( -3 \right)-0+{{c}_{2}} \right)=k \\ & \Rightarrow -8\left( -6+{{c}_{2}} \right)=k \\ & \Rightarrow 48-8{{c}_{2}}=k.......\left( 3 \right) \\ \end{align}
We subtract respective sides of equation (3) from equation (2) to have;
\begin{align} & -36+9{{c}_{2}}=0 \\ & \Rightarrow {{c}_{2}}=\dfrac{36}{9}=4 \\ \end{align}
We put ${{c}_{2}}$ in equation (2) to have;
\begin{align} & 12+4=k \\ & \Rightarrow k=16 \\ \end{align}
We put ${{c}_{2}}=4,k=16$ in equation (1) and find required equation of rectangular hyperbola
$\left( x+2y-5 \right)\left( 2x-y+4 \right)=16$

Note:
We note that an asymptote of curve is a line that meets the curves at infinites. The equation of standard hyperbola is given as $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and the equation of asymptotes of the standard hyperbola is given as $y=\pm \dfrac{b}{a}x$ . The asymptotes of any hyperbola may intersect each other at any angle but asymptotes of a rectangular hyperbola are perpendicular to each other. The standard equation of rectangular hyperbola is $xy={{c}^{2}}$ whose equation of asymptotes are $y=\pm x$ .