$PQ$ is a double ordinate of the parabola $y^2 = 4ax$. What is the locus of the point of intersection of the normals at $P$ and $Q$?

The slopes of the common tangents to the parabola $(x - 1)^2 = 4(y - 2)$ and the ellipse $\frac{(x - 1)^2}{1} + \frac{(y - 2)^2}{2} = 1$ are $m_1$ and $m_2$. Then,$m_1^2 + m_2^2$ is equal to:

Number of intersecting points of the conics $4x^{2} + 9y^{2} = 1$ and $4x^{2} + y^{2} = 4$ is

The triangle $PQR$ of area $A$ is inscribed in the parabola $y^2 = 4ax$ such that the vertex $P$ lies at the vertex of the parabola and the base $QR$ is a focal chord. The modulus of the difference of the ordinates of the points $Q$ and $R$ is:

The value of $b^2$ such that the foci of the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$ and the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ coincide is

$A$ quadratic polynomial $y = f(x)$ with constant term $3$ neither touches nor intersects the $x$-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 vertex of the quadratic polynomial is: