It is known that P is a point on the right branch of hyperbola x2 / 16-y2 / 9 = 1, and F1 and F2 are the left and right focal points respectively. If | Pf1 |: | PF2 | = 3:2, calculate the coordinates of point P!

It is known that P is a point on the right branch of hyperbola x2 / 16-y2 / 9 = 1, and F1 and F2 are the left and right focal points respectively. If | Pf1 |: | PF2 | = 3:2, calculate the coordinates of point P!

x2/16-y2/9=1 → 9x^2-16y^2=144
a=4 b=3
c=5
F1(-5,0) F2=(5,0)
Let the coordinates of point p be (x, y)
Vector f1p = (x + 5,0) vector F2P = (x-5,0)
|F1P|=√[(x+5)^2 +y^2 ]
|F2P|=√[(x-5)^2 +y^2 ]
|PF1|:|PF2|=3:2
√[(x+5)^2 +y^2 ]=3/2 √[(x-5)^2 +y^2 ]
The results are as follows
x^2 +y^2 -26x+250=0
Y ^ 2 = (9x ^ 2 - 144) / 16 is substituted into the above formula
Calculate X and then y