Determine the length of latus rectum of the ellipse $4 {(x - 2 y + 1)}^{2} + 9 {(2 x + y + 2)}^{2} = 25$

\frac{10}{9}

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\frac{17}{9}

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\frac{20}{9}

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\frac{23}{9}

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Solution

The correct option is B $\frac{20}{9}$
Put $x - 2 y + 1 = X, 2 x + y + 2 = Y$
Thus ellipse becomes $4 X^{2} + 9 Y^{2} = 25$
$\Rightarrow \frac{x^{2}}{{(5 / 2)}^{2}} + \frac{Y^{2}}{{(5 / 3)}^{2}} = 1$
$\Rightarrow a = \frac{5}{2}, b = \frac{5}{3}$
Centre is,
$X = 0, Y = 0$ $∴ C (- 1, 0)$
$e^{2} = 1 - \frac{b^{2}}{a^{2}} = 1 - \frac{4}{9} ∴ e = \frac{\sqrt{5}}{3}$
$L . R . = \frac{2 b^{2}}{a} = 2. \frac{25}{9} . \frac{2}{5} = \frac{20}{9}$

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4 (x - 2 y + 1)^{2} + 9 (2 x + y + 2)^{2} = 25

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