The equation of the ellipse passing through t | vedclass

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(D) The center of the ellipse is the midpoint of the foci: $\left( \frac{1+3}{2}, 0 \right) = (2, 0)$.Distance between foci is $2ae = 2$,so $ae = 1$,w

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$3x^2 + 4y^2 = 12x$

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(D) The center of the ellipse is the midpoint of the foci: $\left( \frac{1+3}{2}, 0 \right) = (2, 0)$.Distance between foci is $2ae = 2$,so $ae = 1$,which implies $a^2e^2 = 1$.Using the relation $b^2 = a^2(1 - e^2) = a^2 - a^2e^2$,we get $b^2 = a^2 - 1$.The equation of the ellipse is $\frac{(x-2)^2}{a^2} + \frac{y^2}{b^2} = 1$.Since it passes through the origin $(0, 0)$,we have $\frac{(0-2)^2}{a^2} + \frac{0^2}{b^2} = 1$,which gives $\frac{4}{a^2} = 1$,so $a^2 = 4$.Then $b^2 = 4 - 1 = 3$.The equation is $\frac{(x-2)^2}{4} + \frac{y^2}{3} = 1$.Multiplying by $12$,we get $3(x-2)^2 + 4y^2 = 12$,which simplifies to $3(x^2 - 4x + 4) + 4y^2 = 12$,or $3x^2 + 4y^2 = 12x$.

The equation of the ellipse passing through the origin and having foci at $(1, 0)$ and $(3, 0)$ is .....

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