Let $f(x)=x^2+9, g(x)=\frac{x}{x-9}$ and $\mathrm{a}=\mathrm{fog}(10), \mathrm{b}=\operatorname{gof}(3)$. If $\mathrm{e}$ and $1$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^2}{a}+\frac{y^2}{b}=1$, then $8 e^2+1^2$ is equal to.

Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis, respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S, E_1$ ad $E_2$ at $P, Q$ and $R$, respectively. Suppose that $P Q=P R=\frac{2 \sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$, respectively, then the correct expression$(s)$ is(are)

$(A)$ $e_1^2+e_2^2=\frac{43}{40}$

$(B)$ $e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$

$(C)$ $\left|e_1^2-e_2^2\right|=\frac{5}{8}$

$(D)$ $e_1 e_2=\frac{\sqrt{3}}{4}$

Point $'O' $ is the centre of the ellipse with major axis $AB$ $ \&$ minor axis $CD$. Point $F$ is one focus of the ellipse. If $OF = 6 $  $ \&$  the diameter of the inscribed circle of triangle $OCF$  is $2, $ then the product $ (AB)\,(CD) $ is equal to

Find the equation for the ellipse that satisfies the given conditions: Centre at $(0,\,0),$ major axis on the $y-$ axis and passes through the points $(3,\,2)$ and $(1,\,6)$

An ellipse $\frac{\left(x-x_0\right)^2}{a^2}+\frac{\left(y-y_0\right)^2}{b^2}=1$, $a > b$, is tangent to both $x$ and $y$ axes and is placed in the first quadrant. Let $F_1$ and $F_2$ be two foci of the ellipse and $O$ be the origin with $OF _1 < OF _2$. Suppose the triangle $OF _1 F _2$ is an isosceles triangle with $\angle OF _1 F _2=120^{\circ}$. Then the eccentricity of the ellipse is

If the normal at one end of the latus rectum of an ellipse  $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ passes through one end of the minor axis then :