logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

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

Can you answer this question?
 
 

1 Answer

0 votes
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.
answered Jun 24, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...