The graph of the conic $x^2-(y-1)^2=1$ has one tangent line with positive slope that passes through the origin. If the point of tangency is $(a, b)$,then find the value of $\sin^{-1}\left(\frac{a}{b}\right)$.

The equations of the pairs of opposite sides of a parallelogram are $x^2 - 5x + 6 = 0$ and $y^2 - 6y + 5 = 0$. The equations of its diagonals are:

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(b>a)$ and the parabola $y^2=4ax$ intersect at right angles. If $e$ is the eccentricity of the ellipse,then $2e^2=$

$A$ 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:

Let $PQ$ be a focal chord of the parabola $y^{2}=4x$ such that it subtends an angle of $\frac{\pi}{2}$ at the point $(3, 0)$. Let the line segment $PQ$ be also a focal chord of the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}$. If $e$ is the eccentricity of the ellipse $E$,then the value of $\frac{1}{e^{2}}$ is equal to

$A$ tangent is drawn to the parabola $y^2 = 4x$ at the point $P(t^2, 2t)$,where the abscissa $t^2$ lies in the interval $[1, 4]$. The maximum possible area of the triangle formed by the tangent at $P$,the ordinate of the point $P$,and the $x$-axis is equal to