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  • The length of latus rectum of the ellipse <img alt="\mathrm{4(x-2 y+1)^2+9(2 x+y+2)^2=25}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B4%28x-2%20y&plus;1%29%5E2&plus;9%

The length of latus rectum of the ellipse \mathrm{4(x-2 y+1)^2+9(2 x+y+2)^2=25} is

Option: 1

20 / 9


Option: 2

10 / 9


Option: 3

40 / 9


Option: 4

20 / 16


Answers (1)

best_answer

\mathrm{\begin{aligned} & \text { Put } x-2 y+1=X, 2 x+y+2=Y \\\\ & 4 X^2+9 Y^2=25 \\\\ & \frac{X^2}{(5 / 2)^2}+\frac{Y^2}{(5 / 3)^2}=1 \\\\ & a=5 / 2, b=5 / 3 \\\\ & \text { Centre is } X=0, Y=0 \\\\ & \therefore \quad C \text { is }(-1,0) \\\\ & e^2=1-\frac{b^2}{a^2}=1-\frac{4}{9} . \quad \therefore e=\frac{\sqrt{5}}{3} \\\\ & \text { L.R. }=\frac{2 b^2}{a}=2 \cdot \frac{25}{9} \cdot \frac{2}{5}=\frac{20}{9} \end{aligned}}

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