If (`a x^2+c)y+(a x^2+c )=0a n dx` is a rational function of `ya n da c` is negative, then `a c^(prime)+c^(prime)c=0` b. `a//a '=c//c '` c. `a^2+c^2=a^('2)+c^('2)` d. `a a^(prime)+cc^(prime)=1`

Options

A. `ac' + a'c = 0`
B. `a//a' = c//c'`
C. `a^(2) + c^(2) = a'^(2) + c'^(2)`
D. `aa' + cc' = 1`

Answer

2

Explanation

Given, `(ax^(2) + c)y + (a'x^(2) + c') = 0`
or `x^(2)(ay + a') + (cy + c') = 0`.
If x is rational, then the discriminant of the above equation must be a perfect square. Hence,
`0 - 4 (ay + a') (cy + c')` must be a perfect square
`rArr -acy^(2) - (ac' + a'c)y - a'c'` must be a perfect square
`rArr (ac' + a'c)^(2) - 4ac a' c' = 0" " [because D = 0]`
`rArr (ac' - a'c)^(2) = 0`
`rArr ac' = a' c`
`rArr (a)/(a') = (c)/(c')`