Question

P(x,y) is any point on ellipse `16x^(2)+25y^(2) =400`. If `F_(1)-=(3,0) and F_(2) -=(-3,0)`, then prove that `PF_(1)+PF_(2)` is constant and find its value .

Answer 1

We have, ellipse
`16x^(2)+25y^(2)=400`
or `(x^(2))/(25)+(y^(2))/(16)=1`
`:. a^(2)e^(2)=a^(2)-b^(2)=25-9`
`:. ae=3`
So, foci, are `(+-ae,0) -=(+-3,0)`
Since `F_(1) and F_(2)` are the foci of the ellipse.
Since the sum of the focal distances of a variable point P on an ellipse is equal to its major axis,
`PF_(1)+PF_(2)=2a=10`