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.