Courses for Kids
Free study material
Offline Centres
Store Icon

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
Total views: 386.4k
Views today: 11.86k
386.4k+ views
 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) $ .\[\]

Complete step by step answer:
We are given the equation one asymptote as
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
  & 2\cdot x-1\cdot y+{{c}_{2}}=0 \\
 & \Rightarrow 2x-y+{{c}_{2}}=0 \\
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;
  & \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) \\
We put $ \left( -3,0 \right) $ in equation of rectangular hyperbola (1) and have;
  & \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) \\
We subtract respective sides of equation (3) from equation (2) to have;
  & -36+9{{c}_{2}}=0 \\
 & \Rightarrow {{c}_{2}}=\dfrac{36}{9}=4 \\
We put $ {{c}_{2}} $ in equation (2) to have;
  & 12+4=k \\
 & \Rightarrow k=16 \\
We put $ {{c}_{2}}=4,k=16 $ in equation (1) and find required equation of rectangular hyperbola
\[\left( x+2y-5 \right)\left( 2x-y+4 \right)=16\]

seo images

 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 $ .