Question 2 Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola y^2/9 - x^2/27 = 1
Class X1 - Maths -Conic Sections Page 262
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y²/ 9 - x²/ 27 = 1
y²/(3)² - x²/( 3√3)² = 1 --------(1)
concept : if equation of hyperbola is in the form of y²/a² - x²/b² = 1 -----------(2) then,
major axis is along Y-axis due to coefficient of y² is positive .
vertices ( 0, ± a)
foci ( 0, ± c) where, c² = a² + b²
eccentricity ( e ) = c/a
latusrectum = 2b²/a
now, compare equations (1) and (2) .
a = 3. and. b = 3√3 .
then, c = √{a² + b² } = √{3² + 3√3²} = 6
hence, vertices ( 0, ± a) = ( 0, ± 3)
foci ( 0, ± c) = ( 0, ± 6)
eccentricity ( e ) = c/a = 6/3 = 2
latusrectum = 2b²/a = 2 × 27/3 = 18
y²/(3)² - x²/( 3√3)² = 1 --------(1)
concept : if equation of hyperbola is in the form of y²/a² - x²/b² = 1 -----------(2) then,
major axis is along Y-axis due to coefficient of y² is positive .
vertices ( 0, ± a)
foci ( 0, ± c) where, c² = a² + b²
eccentricity ( e ) = c/a
latusrectum = 2b²/a
now, compare equations (1) and (2) .
a = 3. and. b = 3√3 .
then, c = √{a² + b² } = √{3² + 3√3²} = 6
hence, vertices ( 0, ± a) = ( 0, ± 3)
foci ( 0, ± c) = ( 0, ± 6)
eccentricity ( e ) = c/a = 6/3 = 2
latusrectum = 2b²/a = 2 × 27/3 = 18
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