If the curves ax^2 +4xy+2y^2 +x+y+5=0 and ax^2+6xy+5y^2+2x+3y+8=0 intersect at four con-cyclic points then the value of a is

As we know that, any second degree curve that passes through the intersection of the given curves is ax2 + 4xy + 2y2 + x + y + 5 + λ ( ax2 + 6xy + 5y2 +2x + 3y + 8 ) = 0

So, if it is a circle, then coefficient of x2 = coefficient of y2 and the coefficient of xy = 0

a(1+ λ) = 2 + 5λ and 4 + 6λ = 0

So, a = (2 + 5λ) /(1+λ) and λ = -4/6

Hence, the value of λ is -⅔

Now, substitute λ = -⅔ in a = (2 + 5λ) /(1+λ), we get

a = (2 + 5(-⅔)) /(1+(-⅔))

Now, simplify the above equation, we get a = -4.

Hence, the value of a is -4.

Was this answer helpful?

 
   

0 (0)

(0)
(1)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question