Find the equation for the ellipse that satisf | vedclass

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Question

(A) Since the centre is at $(0, 0)$ and the major axis is on the $y$-axis,the equation of the ellipse is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2

Vedclass · Accepted Answer

$\frac{x^2}{10} + \frac{y^2}{40} = 1$

Vedclass · Answer

(A) Since the centre is at $(0, 0)$ and the major axis is on the $y$-axis,the equation of the ellipse is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $b > a$.Given that the ellipse passes through $(3, 2)$ and $(1, 6)$,we have:$\frac{3^2}{a^2} + \frac{2^2}{b^2} = 1 \implies \frac{9}{a^2} + \frac{4}{b^2} = 1$  --- $(1)$$\frac{1^2}{a^2} + \frac{6^2}{b^2} = 1 \implies \frac{1}{a^2} + \frac{36}{b^2} = 1$  --- $(2)$Let $u = \frac{1}{a^2}$ and $v = \frac{1}{b^2}$. The equations become:$9u + 4v = 1$  --- $(3)$$u + 36v = 1$  --- $(4)$Multiplying $(4)$ by $9$: $9u + 324v = 9$ --- $(5)$Subtracting $(3)$ from $(5)$: $320v = 8 \implies v = \frac{8}{320} = \frac{1}{40}$.Thus,$b^2 = 40$.Substituting $v = \frac{1}{40}$ into $(4)$: $u + 36(\frac{1}{40}) = 1 \implies u + \frac{9}{10} = 1 \implies u = \frac{1}{10}$.Thus,$a^2 = 10$.The equation of the ellipse is $\frac{x^2}{10} + \frac{y^2}{40} = 1$.

Find the equation for the ellipse that satisfies the given conditions: Centre at $(0, 0)$,major axis on the $y$-axis and passes through the points $(3, 2)$ and $(1, 6)$.

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