The equation of the normal drawn at the point $(\sqrt{2}+1, -1)$ to the ellipse $x^2+2y^2-2x+8y+5=0$ is

Find the equation for the ellipse that satisfies the given conditions: Length of minor axis $16$,foci $(0, \pm 6)$.

An ellipse having foci at $(3, 3)$ and $(-4, 4)$ and passing through the origin has eccentricity equal to

Two sets $A$ and $B$ are defined as follows:
$A = \{ (a,b) \in R \times R : |a - 5| < 1 \text{ and } |b - 5| < 1 \}$
$B = \{ (a,b) \in R \times R : 4(a - 6)^2 + 9(b - 5)^2 \le 36 \}$
Then:

Let each of the two ellipses $E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$ and $E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1, (A < B)$ have eccentricity $\frac{4}{5}$. Let the lengths of the latus recta of $E_1$ and $E_2$ be $\ell_1$ and $\ell_2$,respectively,such that $2\ell_1^2 = 9\ell_2$. If the distance between the foci of $E_1$ is $8$,then the distance between the foci of $E_2$ is:

If $(1, -2)$ is the focus and $x+y-2=0$ is the directrix of the ellipse $17x^2 - 2xy + 17y^2 - 32x + 76y + 86 = 0$,then its eccentricity is