1.

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. 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.



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