Question 8 Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 16x^2 + y^2 = 16
Class X1 - Maths -Conic Sections Page 255
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concept : if equation of ellipse is x²/b² + y²/a²=1, ( b < a )
then, vertices ( 0, ± a)
foci( 0, ± c) where c² = a² - b²
The length of major axis = 2a
the length of minor axis = 2b
eccentricity ( e ) = c/a where, c² = a² - b²
length of latusrectum = 2b²/a
16x² + y² = 16
dividing by 16 from both sides,
16x²/16 + y²/16 = 1
x²/1 + y²/16 = 1
x²/1² + y²/4² = 1 compare this with above standard equation .
a = 4 and b = 1 ,
now, c² = a² - b²
c² = 4² - 1² = 16 - 1 = 15
c = √15
now, Foci ( 0, ± c ) = ( 0, ±√15)
vertices ( 0, ±a ) = ( 0, ± 4)
Length of minor axis = 2b = 2 × 1 = 2
Length of major axis = 2a = 2 × 4 = 8
eccentricity ( e ) = c/a = √15/4
Length of Latusrectum = 2b²/a = 2 × 1/4 =1/2
then, vertices ( 0, ± a)
foci( 0, ± c) where c² = a² - b²
The length of major axis = 2a
the length of minor axis = 2b
eccentricity ( e ) = c/a where, c² = a² - b²
length of latusrectum = 2b²/a
16x² + y² = 16
dividing by 16 from both sides,
16x²/16 + y²/16 = 1
x²/1 + y²/16 = 1
x²/1² + y²/4² = 1 compare this with above standard equation .
a = 4 and b = 1 ,
now, c² = a² - b²
c² = 4² - 1² = 16 - 1 = 15
c = √15
now, Foci ( 0, ± c ) = ( 0, ±√15)
vertices ( 0, ±a ) = ( 0, ± 4)
Length of minor axis = 2b = 2 × 1 = 2
Length of major axis = 2a = 2 × 4 = 8
eccentricity ( e ) = c/a = √15/4
Length of Latusrectum = 2b²/a = 2 × 1/4 =1/2
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