For the ellipse 4(x-2y+1)^2 + 9(2x+y+2)^2 = 2 | vedclass

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(A) The given equation is $4(x-2y+1)^2 + 9(2x+y+2)^2 = 25$. Divide by $25$: $\frac{(x-2y+1)^2}{25/4} + \frac{(2x+y+2)^2}{25/9} = 1$. Let $X = \frac{x-

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equation of major axis is $x-2y+1=0$

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(A) The given equation is $4(x-2y+1)^2 + 9(2x+y+2)^2 = 25$. Divide by $25$: $\frac{(x-2y+1)^2}{25/4} + \frac{(2x+y+2)^2}{25/9} = 1$. Let $X = \frac{x-2y+1}{\sqrt{1^2+(-2)^2}} = \frac{x-2y+1}{\sqrt{5}}$ and $Y = \frac{2x+y+2}{\sqrt{2^2+1^2}} = \frac{2x+y+2}{\sqrt{5}}$. The equation becomes $\frac{5X^2}{25/4} + \frac{5Y^2}{25/9} = 1$,which simplifies to $\frac{X^2}{5/4} + \frac{Y^2}{5/9} = 1$. Here $a^2 = 5/4$ and $b^2 = 5/9$,so $a = \sqrt{5}/2$ and $b = \sqrt{5}/3$. The major axis corresponds to the term with the larger denominator,which is $X=0$,i.e.,$x-2y+1=0$. The eccentricity $e = \sqrt{1 - b^2/a^2} = \sqrt{1 - (5/9)/(5/4)} = \sqrt{1 - 4/9} = \sqrt{5}/3$. The length of the major axis is $2a = 2(\sqrt{5}/2) = \sqrt{5}$. The centre is the intersection of $x-2y+1=0$ and $2x+y+2=0$,which is $(-4/5, -2/5)$.

For the ellipse $4(x-2y+1)^2 + 9(2x+y+2)^2 = 25$,which of the following is true?

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