The locus of the point of intersection of perpendicular tangents to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ 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 line $x \cos \alpha + y \sin \alpha = p$ is normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,then

For an ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ with vertices $A$ and $A'$,a tangent drawn at the point $P$ in the first quadrant meets the $y$-axis at $Q$,and the chord $A'P$ meets the $y$-axis at $M$. If $O$ is the origin,then $OQ^2 - MQ^2$ is equal to:

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

The equation of the normal to the curve $4x^2 + 9y^2 = 36$ at the point where the parametric angle is $\theta = \frac{7\pi}{4}$ is