If `(x-a)costheta+ysintheta=(x-a)cosvarphi+ysintheta=aa n dtan(theta/2)-tan(varphi/2)=2b ,t h e n` `y^2=2a x-(1-b^2)x^2` `tantheta/2=1/x(y+b x)` `y^2=2b x-(1-a^2)x^2` `tanvarphi/2=1/x(y-b x)`

Options

A. `y^(2)=2ax-(1-b^(2))x^(2)`
B. `tan((theta)/(2))=(1)/(x)(y+bx)`
C. `y^(2)=2bx-(1-a^(2))x^(2)`
D. `tan((phi)/(2))=(1)/(x)(y-bx)`

Answer

A, B, D

Explanation

Let `tan(theta//2) = alpha and tan ( phi //2) = beta`, so that `alpha - beta = 2b`.
Also, `cos theta = (1 - tan^(2)(theta//2))/( 1+tan^(2)(theta//2)) = (1-alpha^(2))/(1+alpha^(2))`
and `" " sin theta = ( 2tan(theta//2))/( 1+ tan^(2)( theta//2)) = ( 2alpha )/(1+ alpha^(2))`
Similarly, `cos phi = (1- beta^(2))/(1+beta^(2)) and sin phi = (2beta)/( 1+ beta^(2))`
Therefore, we have from the given realtions.
`" " (x-a) (1-alpha ^(2))/(1+alpha^(2)) + y((2alpha)/(1+alpha^(2)) =a`
`rArr x alpha^(2) - 2yalpha + 2a - x =0`
Similarly `xbeta^(2) - 2y beta + 2a -x =0`
We see that `alpha and beta ` are roots of the equation.
`" " xz^(2) - 2yz + 2a -x =0`
So that `alpha + beta = 2y//x and alphabeta = (2a - x)//x`.
Now, from `(alpha + beta)^(2) = ( alpha - beta)^(2) + 4alpha beta`, we get
`rArr ((2y)/(x))^(2) = (2b)^(2) + (4(2a-x))/(x)`
`rArr y^(2) = 2ax - (1-b^(2))x^(2)`
Also, from `alpha + beta = 2y//x and alpha - beta = 2b`, we get
`alpha = y//x + b and beta = y//x - b`
`rArr tan ""(theta)/(2) = (1)/(x)(y+ bx) and tan""(phi)/(2) = (1)/(x) (y-bx)`