Hi,
Your answer for (1) is correct.
The centre of the hyperbola is at the origin. The
hyperbola has vertices at (-4, 0) and (4, 0). Each focus
of the hyperbola is c units from the centre, where
c^2 = 16 + 20 (the sum of the denominators),
from which we obtain c = 6.
Now, the equation of the ellipse is
(x^2 / 6^2) + (y^2 / b^2) = 1, where b is to be
found.
The centre of the ellipse is at the origin. Each focus is
4 units from the centre, where
4^2 = 6^2 - b^2 (the difference of the
denominators),
from which we obtain b^2 = 20.
For (2):
1. Let position vector of P be xi +
yj.
2. Form the vectors PA and PF.
3. Form an equation, given the zero scalar product of
vectors PA and PF.
4. With the equation obtained for the ellipse, solve the
equation in step 3.
Is this a question from IB Maths HL?
Thanks.
Cheers,
Wen Shih
