Q. 115.0( 2 Votes )

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

Answer :

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) (a^{2}>b^{2}).

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

⇒ ae = 3

⇒

⇒ a^{2} - b^{2} = 9 ..... - - - (2)

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

⇒

⇒

⇒ 16b^{2} + a^{2} = a^{2}b^{2}

From (2),

⇒ 16(a^{2} - 9) + a^{2} = a^{2}(a^{2} - 9)

⇒ 16a^{2} - 144 + a^{2} = a^{4} - 9a^{2}

⇒ a^{4} - 26a^{2} + 144 = 0

⇒ a^{4} - 18a^{2} - 8a^{2} + 144 = 0

⇒ a^{2}(a^{2} - 18) - 8(a^{2} - 18) = 0

⇒ (a^{2} - 8)(a^{2} - 18) = 0

⇒ a^{2} - 8 = 0 (or) a^{2} - 18 = 0

⇒ a^{2} = 8 (or) a^{2} = 18

⇒ b^{2} = 18 - 9(since b^{2}>0)

⇒ b^{2} = 9.

The equation of the ellipse is

⇒

⇒

⇒ x^{2} + 2y^{2} = 18

∴ The equation of the ellipse is x^{2} + 2y^{2} = 18.

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