An ellipse passes through the point $(-3, 1)$ and its eccentricity is $\sqrt {\frac{2}{5}} $. The equation of the ellipse is

Consider the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. Let $S(p, q)$ be a point in the tirst quadrant such that $\frac{p^2}{9}+\frac{q^2}{4}>1$. I wo tangents are drawn from $S$ to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point $T$ in the fourth quadrant. Let $R$ be the vertex of the ellipse with positive $x$-coordinate and $O$ be the center of the ellipse. If the area of the triangle $\triangle O R T$ is $\frac{3}{2}$, then which of the following options is correct?

The equation of the ellipse whose one of the vertices is $(0,7)$ and the corresponding directrix is $y = 12$, is

The point $(4, -3)$ with respect to the ellipse $4{x^2} + 5{y^2} = 1$

Two sets $A$ and $B$ are as under:

$A = \{ \left( {a,b} \right) \in R \times R:\left| {a - 5} \right| < 1 \,\,and\,\,\left| {b - 5} \right| < 1\} $; $B = \left\{ {\left( {a,b} \right) \in R \times R:4{{\left( {a - 6} \right)}^2} + 9{{\left( {b - 5} \right)}^2} \le 36} \right\}$ then : . . . . .

If the tangent to the parabola $y^2 = x$ at a point $\left( {\alpha ,\beta } \right)\,,\,\left( {\beta  > 0} \right)$ is also a tangent to the ellipse, $x^2 + 2y^2 = 1$, then $a$ is equal to