Question

find equtation of ellipse(i) eccentricity
2/3
and length of latus-rectum =5​

Answer 1

Step-by-step explanation:

The Answer is simply from there eq's

# x²/a² + y²/b² = 1 ------>> Eqñ of Ellipse

# Length of LR = 2b²/a ............(2)

# Relñ. b/w a,b &e =

b²=a²(1-e²)...........(3)

===> Thus, we just need A & B

From eqñ(2)

LR =5= 2b²/a

=> b² = 5a/2

==> Put in Eqñ (3)

5a/2 = a²(1-2²/3²)

=> 5/2 = a(5/9)

=> 1/2 = a/9

=> a= 9/2

Put a=9/2 and e=2/3

b² = 81/4 (5/9)

b² = 45/4

Thus, Eqñ of Ellipse

=> x²/a² + y²/b² = 1

=> 4x²/81 + 4y²/45 = 1

or x²/81 + y²/45 = 1/4