Find the equation of an ellipse whose foci are at(± 3, 0) and which passes through (4, 1).

Given that we need to find the equation of the ellipse whose foci are at (±4,0) and passes through (4,1).

Let us assume the equation of the ellipse is - - - - (1) (a2>b2).

We know that foci are (±ae,0) and eccentricity of the ellipse is

ae = 3

a2 - b2 = 9 ..... - - - (2)

Substituting the point (4,1) in (1) we get,

16b2 + a2 = a2b2

From (2),

16(a2 - 9) + a2 = a2(a2 - 9)

16a2 - 144 + a2 = a4 - 9a2

a4 - 26a2 + 144 = 0

a4 - 18a2 - 8a2 + 144 = 0

a2(a2 - 18) - 8(a2 - 18) = 0

(a2 - 8)(a2 - 18) = 0

a2 - 8 = 0 (or) a2 - 18 = 0

a2 = 8 (or) a2 = 18

b2 = 18 - 9(since b2>0)

b2 = 9.

The equation of the ellipse is

x2 + 2y2 = 18

The equation of the ellipse is x2 + 2y2 = 18.

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