Equation of the normal to the hyperbola $\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{16}} = 1$ perpendicular to the line $2x + y = 1$ is

Consider a branch of the hyperbola $x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0$ with vertex at the point $A$. Let $B$ be one of the end points of its latus rectum. If $\mathrm{C}$ is the focus of the hyperbola nearest to the point $\mathrm{A}$, then the area of the triangle $\mathrm{ABC}$ is

$P(6, 3)$ is a point on the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ . If the normal at point $P$ intersect the $x-$ axis at $(10, 0)$ , then the eccentricity of the hyperbola is

The graph of the conic $ x^2 - (y - 1)^2 = 1$  has one tangent line with positive slope that passes through the origin. the point of tangency being $(a, b). $ Then  Length of the latus rectum of the conic is

Let the focal chord of the parabola $P: y^{2}=4 x$ along the line $L: y=m x+c, m>0$ meet the parabola at the points $M$ and $N$. Let the line $L$ be a tangent to the hyperbola $H : x ^{2}- y ^{2}=4$. If $O$ is the vertex of $P$ and $F$ is the focus of $H$ on the positive $x$-axis, then the area of the quadrilateral $OMFN$ is.

Eccentricity of conjugate hyperbola of $16x^2 - 9y^2 - 32x - 36y - 164 = 0$ will be-