The equation $\frac{x^{2}}{2-\lambda}-\frac{y^{2}}{\lambda-5}-1=0$ represents an ellipse,if

The equations of the latus rectum of the ellipse $9x^2 + 25y^2 - 36x + 50y - 164 = 0$ are

If $P$ lies in the first quadrant on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b$),and the tangent and normal drawn at $P$ meet the major axis at points $T$ and $N$ respectively,then the value of $\frac{(\left| F_2N \right| + \left| F_1N \right|)(\left| F_2T \right| - \left| F_1T \right|)}{(\left| F_2N \right| - \left| F_1N \right|)(\left| F_2T \right| + \left| F_1T \right|)}$ is equal to (where $F_1$ and $F_2$ are the foci $(ae, 0)$ and $(-ae, 0)$ respectively).

$S^{\prime}$ is the focus of the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1, (b < 5)$ lying on the negative $X$-axis and $P(\theta)$ is a point on this ellipse. If the distance between the foci of this ellipse is $8$ and $S^{\prime}P = 7$,then $\theta =$

The locus of the point of intersection of the perpendicular tangents to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is

If a tangent having slope $m = \frac{1}{3}$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is a normal to the circle $(x + 1)^2 + (y + 1)^2 = 1$,then $a^2$ lies in the interval: