Find the axis of the parabola $9y^2 - 16x - 12y - 57 = 0$.

Let a line $y=mx$ $(m>0)$ intersect the parabola $y^{2}=x$ at a point $P$,other than the origin. Let the tangent to it at $P$ meet the $x$-axis at the point $Q$. If $\text{area}(\Delta OPQ)=4$ sq. units,then $m$ is equal to

The tangents at the points $A(1, 3)$ and $B(1, -1)$ on the parabola $y^{2} - 2x - 2y = 1$ meet at the point $P$. Then the area (in unit$^{2}$) of the triangle $PAB$ is:

Let $L_1, L_2$ be the lines passing through the point $P(0,1)$ and touching the parabola $9x^2+12x+18y-14=0$. Let $Q$ and $R$ be the points on the lines $L_1$ and $L_2$ such that the $\triangle PQR$ is an isosceles triangle with base $QR$. If the slopes of the lines $QR$ are $m_1$ and $m_2$,then $16(m_1^2+m_2^2)$ is equal to ..............

The two parabolas $y^2 = 4x$ and $x^2 = 4y$ intersect at a point $P$,whose abscissa is not zero,such that

Let chord $PQ$ of length $3\sqrt{13}$ of the parabola $y^2 = 12x$ be such that the ordinates of points $P$ and $Q$ are in the ratio $1:2$. If the chord $PQ$ subtends an angle $\alpha$ at the focus of the parabola,then $\sin \alpha$ is equal to: