A rectangular Hyperbola is drawn thru the point of intersection of the circle x2+y2+2gx+2fy+c=0
& the pair of straight lines ax2 + by2+ 2hxy=0.
If the hyperbola cuts x axis at A & B & the y-axis at C & D . The equation of the line joining the mid points of AB & CD is
x/g - y/f = ....
(A) ab/(a-b)
(B) (a-b)/(a+b)
(C) ab/(a+b)
(D) (A+b)/(a-b)
29Assume the eqn of the rectangular hyperbola to be x2 + y 2 + 2gx + 2fy +c + λ(ax2 + by2 +2hxy ) = 0...
then find out the mid point of A and B by putting y = 0
u will get \frac{x_{1}+x_{2}}{2} = -\frac{2}{\lambda a+ 1}
\frac{y_{1}+y_{2}}{2} = -\frac{2}{\lambda b+ 1}
Now for a rectangular hyperbola ( λa + 1) = - (λb + 1)
so λ = \frac{-2}{a+b}
now substituting the value of λ in the eqn x/g - y/f = 2(a+b)/(a-b)
4@Govind I dint under stand this part ""for a rectangular hyperbola ( λa + 1) = - (λb + 1) ""
Y is this condition true??
29eqn of hyperbola = \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1
For it to be a rectangular hyperbola a^{2} = b^{2}
So from there i got this condition from there....well i have verified it with one question in which eqn of hyperbola was given as 2nd degree curve like abv...so it seems this condition is true...
4Ya fine agreed that condition is correct
but
That means coeff of x = coeff of y na?
How u got this " ( λa + 1) = - (λb + 1) "
29Assume the eqn of the rectangular hyperbola to be x2 + y2 + 2gx + 2fy +c + λ(ax2 + by2 +2hxy ) = 0....
from here u will get the condition
4Ok ya I got it
I think u're ans is correct ( I dint check the final calculation part)
But u're method is nice n simple
Thanks
