Question

find the coordinates of the foci vertices the length of major axis of this equation 36x square +y square=144​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Foci ( 0, +/-c) = ( 0, +/- 4√2)

Step-by-step explanation:

if equation of ellipse is x²/b² + y²/a² = 1(b<a) then,

vertices (0, +/-a)

foci ( 0, +/-c) where, c² = a²- b²

Length of minor axis = 2b

length of major axis = 2a

eccentricity ( e ) = c/a

length of latusrectum = = 2b²/a

Here,

36x² + 4y² = 144

36x² /144 + 4y²/144 = 144/144

x²/4 + y²/36 = 1

x²/2² + y²/6² = 1 compare this equation with above standard equation of ellipse.

we get, a = 6 and b = 2

now,c² = a² - b²

c² = 6² - 2² = 36 - 4 = 32

c = 4√2

Hence, vertices ( 0, +/-a) = ( 0, +/-6)

Foci ( 0, +/-c) = ( 0, +/- 4√2)

Length of minor axis = 2b = 2 x 2 = 2

Length of major axis = 2a = 2 x 6 = 12

eccentricity ( e ) = c/a = 4√2/6 = 2√2/3

Length of latusrectum = 2b²/a = 2 x 4/6 = 4/3

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