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 is $(a, b)$. Then the eccentricity of the conic is:

The equations of the tangents to the hyperbola $4x^2 - y^2 = 12$ are $y = 4x + c_1$ and $y = 4x + c_2$. Then $|c_1 - c_2|$ is equal to -

The line $2x + \sqrt{6}y = 2$ is a tangent to the curve $x^2 - 2y^2 = 4$. The point of contact is

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ where $\theta + \phi = \frac{\pi}{2}$ be two points on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. If $(h, k)$ is the point of intersection of the normals drawn at $P$ and $Q$,then $k=$

The equation of the directrices of the hyperbola $3x^{2}-3y^{2}-18x+12y+2=0$ is

The locus of the midpoints of the chord of the circle $x^{2}+y^{2}=25$ which is tangent to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ is