Answer :

Given:

16x^{2} + 25y^{2} = 400

Divide by 400 to both the sides, we get

…(i)

Since, 25 > 4

So, above equation is of the form,

…(ii)

Comparing eq. (i) and (ii), we get

a^{2} = 25 and b^{2} = 4

⇒ a = √25 and b = √4

⇒ a = 5 and b = 2

(i) __To find__: Length of major axes

Clearly, a > b, therefore the major axes of the ellipse is along x axes.

∴Length of major axes = 2a

= 2 × 5

= 10 units

(ii) __To find__: Coordinates of the Vertices

Clearly, a > b

∴ Coordinate of vertices = (a, 0) and (-a, 0)

= (5, 0) and (-5, 0)

(iii) __To find__: Coordinates of the foci

We know that,

Coordinates of foci = (±c, 0) where c^{2} = a^{2} – b^{2}

So, firstly we find the value of c

c^{2} = a^{2} – b^{2}

= 25 – 4

c^{2} = 21

c = √21 …(I)

∴ Coordinates of foci = (±√21, 0)

(iv) __To find__: Eccentricity

We know that,

[from (I)]

(v) __To find__: Length of the Latus Rectum

We know that,

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