A(-a,0); B(a,0) are fixed points. C is a point which divides AB in a constant ratio tan . If AC & CB subtend equal angles at P, prove that the equation of the locus of P is x² + y² + 2ax cosec 2 + a² = 0.

Dear,

Let P be (x,y).

in triangle APC , LET <APC = A  and < PCA = B

 hence ,<BPC = A    and <PCB = (180-B)

Apply sin  Rule  in both triangles:

(Sin A / AC ) = (Sin B/ AP)  ..................eq(1)

and (Sin A / CB ) = (Sin B/ PB).................eq(2)

from (1) and (2) , equate Sin A/Sin B:

(AC / AP) = (CB / PB)

or (AC / CB) = (AP / PB)

or tan = (AP / PB)                             since (AC/CB) = tan

or PB tan = AP

or PB² tan² = AP²

or [(x-a)²+y² ] tan² = [(x+a)²+y² ]

or x²(tan² -1)+y² (tan² -1) + a² (tan² -1) - 2ax (tan² +1)=0

or x²+y²+a² - 2ax (tan² +1)/(tan² -1)=0  

or x²+y²+a² + 2ax (tan² +1)/(1-tan²)=0  

or x²+y²+a² + 2ax (Sin² +Cos²)/(Cos²-Sin²)=0   

or x²+y²+a² + 2ax (1)/(Cos2)=0  

or x²+y²+a² + 2ax Sec2=0

please check your question it may be sec instead of cosec.