The equation of the ellipse whose centre is a | vedclass

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(B) The standard equation of an ellipse with its centre at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the ellipse passes through the

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$3x^2 + 5y^2 = 32$

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(B) The standard equation of an ellipse with its centre at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the ellipse passes through the points $(-3, 1)$ and $(2, -2)$,we substitute these coordinates into the equation:For $(-3, 1)$: $\frac{(-3)^2}{a^2} + \frac{1^2}{b^2} = 1 \implies \frac{9}{a^2} + \frac{1}{b^2} = 1$  (Equation $1$)For $(2, -2)$: $\frac{2^2}{a^2} + \frac{(-2)^2}{b^2} = 1 \implies \frac{4}{a^2} + \frac{4}{b^2} = 1 \implies \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$ (Equation $2$)Subtracting Equation $2$ from Equation $1$:$(\frac{9}{a^2} + \frac{1}{b^2}) - (\frac{1}{a^2} + \frac{1}{b^2}) = 1 - \frac{1}{4}$$\frac{8}{a^2} = \frac{3}{4} \implies a^2 = \frac{32}{3}$Substituting $a^2$ into Equation $2$:$\frac{1}{32/3} + \frac{1}{b^2} = \frac{1}{4} \implies \frac{3}{32} + \frac{1}{b^2} = \frac{8}{32}$$\frac{1}{b^2} = \frac{5}{32} \implies b^2 = \frac{32}{5}$Substituting $a^2$ and $b^2$ into the standard equation:$\frac{x^2}{32/3} + \frac{y^2}{32/5} = 1 \implies \frac{3x^2}{32} + \frac{5y^2}{32} = 1$$3x^2 + 5y^2 = 32$.

The equation of the ellipse whose centre is at the origin and which passes through the points $(-3, 1)$ and $(2, -2)$ is

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