If $e$ and $e^{\prime}$ are the eccentricities of the ellipse $5x^2 + 9y^2 = 45$ and the hyperbola $5x^2 - 4y^2 = 45$ respectively,then $ee^{\prime}$ is equal to

The ellipse $4x^2 + 9y^2 = 36$ and the hyperbola $4x^2 - y^2 = 4$ have the same foci and they intersect at right angles. Then,the equation of the circle passing through the points of intersection of the two conics is:

If the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ coincide with the foci of the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$,then $b^2$ is equal to

$A$ quadratic polynomial $y = f(x)$ with absolute term $3$ neither touches nor intersects the abscissa axis and is symmetric about the line $x = 1$. The coefficient of the leading term of the polynomial is unity. $A$ point $A(x_1, y_1)$ with abscissa $x_1 = 1$ and a point $B(x_2, y_2)$ with ordinate $y_2 = 11$ are given in a Cartesian rectangular system of coordinates $OXY$ in the first quadrant on the curve $y = f(x)$,where $O$ is the origin. The scalar product of the vectors $\vec{OA}$ and $\vec{OB}$ is:

The locus of the midpoints of the chords of the hyperbola $x^2 - y^2 = a^2$ which are tangents to the parabola $x^2 = 4by$ will be -

$A$ circle is described whose centre is the vertex and whose diameter is three-quarters of the latus rectum of the parabola $y^2 = 4ax$. If $PQ$ is the common chord of the circle and the parabola and $L_1L_2$ is the latus rectum,then the area of the trapezium $PL_1L_2Q$ is: