The equation of the circle which cuts all the three circles $4(x-1)^2+4(y-1)^2=1,4(x+1)^2+4(y-1)^2$ and $4(x+1)^2+4(y+1)^2=1$ orthogonally is

If the normal chord drawn at the point $\left(\frac{15}{2}, \frac{15}{\sqrt{2}}\right)$ to the parabola $y^2=15 x$ subtends an angle $\theta$ at the vertex of the parabola, then $\sin \frac{\theta}{3}+\cos \frac{2 \theta}{3}-\sec \frac{4 \theta}{3}=$

If a tangent having slope $\frac{1}{3}$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$ is a normal to the circle $(x+1)^2+(y+1)^2=1$, then $a^2$ lies in the interval

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 $K=$