​ In the figure S and S’ are foci of the ellipse, x^2/25 + y^2/16 and P is a viable point on the ellipse. ​

Question

In the figure S and S’ are foci of the ellipse, \(\frac{x^2}{25}+ \frac{y^2}{16} =1\) and P is a viable point on the ellipse.

1. Find the co-ordinate of foci. 

2. Find the distance between S and S1.

3. What is the maximum area of the triangle PSS’.

Answer 1

1. Since 25 > 16 the standard equation of the ellipse is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) 

⇒ a² = 25; b² = 16

c² = a² – b² = 25 – 16 = 9 ⇒ c = 3

Coordinate of foci are (±3, 0).

2. Distance between the two focus is 3 + 3 = 6.

3. Maximum area is attained when the point P reaches the point the ellipse meet the y
axis. Then area is = \(\frac{1}{2}\) × 6 × 4 = 12.