If the extremities of the latus rectum having positive ordinate of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ lie on the parabola $x^2 + 2ay - 4 = 0$,then the points $(a, b)$ lie on the curve:

Let $a, b$ and $\lambda$ be positive real numbers. Suppose $P$ is an end point of the latus rectum of the parabola $y^2 = 4 \lambda x$,and suppose the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ passes through the point $P$. If the tangents to the parabola and the ellipse at the point $P$ are perpendicular to each other,then the eccentricity of the ellipse is

The tangent at $P$ to a parabola $y^2 = 4ax$ meets the directrix at $U$ and the latus rectum at $V$. Then $\triangle SUV$ (where $S$ is the focus) :

The line $y=x+5$ touches

If the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ are the same as the foci of the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,then $b^2 = \dots$

If the normal to a parabola $y^2 = 4ax$ at point $P$ meets the curve again at point $Q$,and if $PQ$ and the normal at $Q$ make angles $\alpha$ and $\beta$ respectively with the $x$-axis,then the value of $\tan \alpha (\tan \alpha + \tan \beta)$ is: