The length of the latus-rectum of the ellipse, whose foci are $(2,5)$ and $(2,-3)$ and eccentricity is $\frac{4}{5}$, is

<p>To find the length of the latus rectum of the ellipse, we first recognize that the foci of the ellipse are given as $ F_1:(2,5) $ and $ F_2:(2,-3) $. This indicates that the major axis is aligned along the $ y $-axis.</p> <p><p><strong>Calculate the distance between the foci:</strong></p> <p>$ F_1F_2 = 8 $</p> <p>The formula involving the distance between the foci and the eccentricity is:</p> <p>$ F_1F_2 = 2be $</p> <p>Here, the eccentricity $ e $ is given as $ \frac{4}{5} $. Thus, we can solve for $ b $:</p> <p>$ 8 = 2b \cdot \frac{4}{5} \implies b = \frac{8}{2 \times \frac{4}{5}} = 5 $</p></p> <p><p><strong>Determine $ a^2 $:</strong></p> <p>Using the relationship between eccentricity, semi-minor axis $ b $, and semi-major axis $ a $:</p> <p>$ e^2 = 1 - \frac{a^2}{b^2} = 1 - \frac{a^2}{25} = \frac{16}{25} $</p> <p>Solving for $ a^2 $:</p> <p>$ 1 - \frac{a^2}{25} = \frac{16}{25} \implies \frac{a^2}{25} = \frac{9}{25} \implies a^2 = 9 $</p> <p>Thus, $ a = 3 $.</p></p> <p><p><strong>Compute the length of the latus rectum:</strong></p> <p>The formula for the length of the latus rectum $ L $ is:</p> <p>$ L = \frac{2a^2}{b} $</p> <p>Substituting the known values:</p> <p>$ L = \frac{2 \times (9)}{5} = \frac{18}{5} $</p></p> <p>Therefore, the length of the latus rectum of the ellipse is $ \frac{18}{5} $.</p>