Let P be a parabola with vertex (2, 3) and directrix 2x + y = 6 ....

Let $Z$ be the foot of perpendicular from vertex to directrix of parabola,

Now, finding $Z$ we get,

$⇒ \dfrac{x - 2}{2} = \dfrac{y - 3}{1} = - \dfrac{\left(4 + 3\right) - 6}{4 + 1}$

$⇒ \dfrac{x - 2}{2} = \dfrac{y - 3}{1} = \dfrac{- 1}{5}$

$⇒ \dfrac{x - 2}{2} = \dfrac{- 1}{5}, \dfrac{y - 3}{1} = \dfrac{- 1}{5}$

$⇒ x = \dfrac{12}{5}, y = \dfrac{16}{5}$

$⇒ Z ≡ \left(\dfrac{12}{5}, \dfrac{16}{5}\right)$

Eccentricity of ellipse is given as $\dfrac{1}{\sqrt{2}}$.

Now, finding ${b}^{2}$ using eccentricity formula we get,

$⇒ {b}^{2} = {a}^{2}\left(1 - {e}^{2}\right) = \dfrac{{a}^{2}}{2}$

So, equation of ellipse will be,

$⇒ \dfrac{144}{25{a}^{2}} + \dfrac{256}{25 \times \dfrac{{a}^{2}}{2}} = 1$ {as $\left(x, y\right) ≡$\left(\dfrac{12}{5}, \dfrac{16}{5}\right)}</p><p>⇒ \dfrac{144}{25{a}^{2}} + \dfrac{512}{25{a}^{2}} = 1$</p><p>$⇒ \dfrac{656}{25{a}^{2}} = 1$</p><p>$⇒ {a}^{2} = \dfrac{656}{25}$</p><p>$⇒ {b}^{2} = \dfrac{328}{25}$</p><p>Length of latus rectum is given by,<br/>$L = \dfrac{2{b}^{2}}{a}$.</p><p>$⇒ L = \dfrac{2 \times \dfrac{328}{25}}{\dfrac{\sqrt{656}}{5}}$</p><p>$⇒ L = \dfrac{\sqrt{656}}{5}$</p><p>$⇒ {L}^{2} = \dfrac{656}{25}