The eccentricity of the ellipse given by the equation $9x^{2} + 16y^{2} = 144$ is

The chord joining two points $\theta_1$ and $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ subtends a right angle at the . . . point. (Given $\tan \theta_1 \tan \theta_2 = -\frac{a^2}{b^2}$)

$S$ and $T$ are the foci of an ellipse and $B$ is the end point of the minor axis. If $\triangle STB$ is an equilateral triangle,the eccentricity of the ellipse is

The product of the lengths of the perpendiculars drawn from the two foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ to the tangent at any point on the ellipse is

Let $E_1 = \frac{x^2}{9} + \frac{y^2}{4} = 1$ and $E_2 = \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be two ellipses and $R$ be a rectangle with sides parallel to the coordinate axes. Let $E_1$ be the inscribed ellipse in $R$ and $E_2$ be the circumscribed ellipse on $R$. If $E_2$ passes through $(0, 4)$,then:

If the normal drawn at the point $P(\frac{\pi}{4})$ on the ellipse $x^2+4y^2-4=0$ meets the ellipse again at $Q(\alpha, \beta)$,then $\alpha=$