The length of the latus-rectum of the ellipse, whose foci are (2,5) and (2,-3) and eccentr…

Question

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

Accepted Answer

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.

Calculate the distance between the foci:

$    F_1F_2 = 8    $

The formula involving the distance between the foci and the eccentricity is:

$    F_1F_2 = 2be    $

Here, the eccentricity $ e $ is given as $ \frac{4}{5} $. Thus, we can solve for $ b $:

$    8 = 2b \cdot \frac{4}{5} \implies b = \frac{8}{2 	imes \frac{4}{5}} = 5    $

Determine $ a^2 $:

Using the relationship between eccentricity, semi-minor axis $ b $, and semi-major axis $ a $:

$    e^2 = 1 - \frac{a^2}{b^2} = 1 - \frac{a^2}{25} = \frac{16}{25}    $

Solving for $ a^2 $:

$    1 - \frac{a^2}{25} = \frac{16}{25} \implies \frac{a^2}{25} = \frac{9}{25} \implies a^2 = 9    $

Thus, $ a = 3 $.

Compute the length of the latus rectum:

The formula for the length of the latus rectum $ L $ is:

$    L = \frac{2a^2}{b}    $

Substituting the known values:

$    L = \frac{2 	imes (9)}{5} = \frac{18}{5}    $

Therefore, the length of the latus rectum of the ellipse is $ \frac{18}{5} $.

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