Consider the ellipse 64x2 16y2 49 Its vertices are 0 plusm

Consider the ellipse 64x^2 + 16y^2 = 49. Its vertices are (0, plusminus A)with A = its foci are (0, plusminus B) with B = its eccentricity is the length of its major axe is the length of its minor axe is

Solution

11.2.1)

64x²+16y²=49

(64x²/49)+(16y²/49)=1

(x²/(49/64))+(y²/(49/16))=1

(x/(7/8))²+(y/(7/4))²=1

a=7/8 , b=7/4

length of major axis =b = 2*(7/4)=7/2

length of minor axis =a = 2*(7/8)=7/4

vertices (0,+-A) ,A=7/4

eccentricity =(1-(a/b)²)

eccentricity =(1-(1/2)²) =(3)/2

foci=(0,-+B)

B=be

B=(7/4)(3)/2

B=(7/8)3