If the length of the latus rectum of the ellipse x ^ 2 + 4 y ^ 2 +...

Given,

${x}^{2} + 4{y}^{2} + 2x + 8y - \lambda = 0$ is a ellipse

Now on rearranging the terms we get,

$⇒ \dfrac{{\left(x + 1\right)}^{2}}{\lambda + 5} + \dfrac{{\left(y + 1\right)}^{2}}{\dfrac{\lambda + 5}{4}} = 1$

We know that length of latus rectum is $\dfrac{2{b}^{2}}{a} = 4$

So, $\dfrac{2\left(\lambda + 5\right)}{4\left(\sqrt{\lambda + 5}\right)} = 4$

$⇒ \dfrac{2\left(\lambda + 5\right)}{4} = 4\left(\sqrt{\lambda + 5}\right)$

$⇒ \left(\lambda + 5\right) = 8\left(\sqrt{\lambda + 5}\right)$

$⇒ {\left(\lambda + 5\right)}^{2} = 64\left(\lambda + 5\right)$

$⇒ {\lambda}^{2} + 25 + 10\lambda = 64\left(\lambda + 5\right)$

$⇒ {\lambda}^{2} - 54\lambda - 295 = 0$

$⇒ \lambda = 59\text{or} - 5$

But $\lambda \ne - 5$

So, $l = 2a = 2\sqrt{\lambda + 5} = 2\sqrt{64} = 16$

$⇒ \lambda + l = 59 + 16 = 75$