For the ellipse 25 x²+9 y²-150 x-90 y+225=0 the eccentricity e is equal to

Step-by-step solution

We have, aligned & 25 x²+9 y²-150 x-90 y+225=0 & 25 (x²-6 x )+9 (y²-10 y )+225=0 & 25 [(x-3)²-9 ]+9 [(y-5)²-25 ]+225=0 & 25(x-3)²-225+9(y-5)²-225+225=0 & 25(x-3)²+9(y-5)²=225 & a=3, b=5 On comparing with (x-h)² a² + (y-k)² b² =1, we have & Eccentricity, e= 1- a² b² = 1- 9 25 = 4 5 aligned Eccentricity, e= 1- a² b² = 1- 9 25 = 4 5