$2 x^{2} + y^{2} - 8 x - 2 y + 1 = 0.$
Find the square of the Latus Rectum for the given ellipse.

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Solution

Given equation $2 x^{2} + y^{2} - 8 x - 2 y + 1 = 0$ .
By simplifying we get $2 {(x - 2)}^{2} + {(y - 1)}^{2} = 8 o r \frac{{(x - 2)}^{2}}{4} + \frac{{(y - 1)}^{2}}{8} = 1$ or
$\frac{X^{2}}{4} + \frac{Y^{2}}{8} = 1$ ,where $X = (x - 2), Y = (y - 1), b^{2} = 8, a^{2} = 4$
Here Y-axis is major axis,now length of L.R is given by $\frac{2 a^{2}}{b} = \frac{8}{\sqrt{8}} = 2 \sqrt{2}$
Square of it is equal to $8$ .

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