If the perimeter of a triangle is 20 and two of its vertices are $(-5,0)$ and $(6,0)$, then the locus of the third vertex is

Let $S$ be the focus of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ lying on the positive $X$ - axis and $P\left(5, y_1\right)$ be point on the hyperbola. Then $S P=$

If $P(\theta)=\left(x_1, \frac{3 \sqrt{5}}{2}\right), 0<\theta<\frac{\pi}{2}$ is a point on the hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$, where $\theta$ is the parameter in its parametric form, then $2 x_1+9 \sin ^2 \theta=$

If $\frac{x^2}{k-\frac{5}{2}}+\frac{y^2}{\frac{7}{3}-k}=1$ ( $k$ is a real number) represents a hyperbola, then the set of all values of $k$ is