R S Aggarwal Solutions for Exercise 1: EXERCISE 5

The equation of the hyperbola whose vertices are $\left(\pm 2,0\right)$ and the eccentricity is $2$ is $\frac{x^2}{k^2}-\frac{y^2}{12}=1$, then find the value of $k^2$.

The equation of the hyperbola whose foci are $\left(\pm \sqrt{5},0\right)$ and the eccentricity is $\sqrt{\frac{5}{3}}$ is $\frac{x^2}{3}-\frac{y^2}{k^2}=1$, then find the value of $k^2$.

The equation of the hyperbola, the length of whose latus rectum is $4$ and the eccentricity is $3$ is $k{x}^2-2y^2=1$, then find the value of $k$.

The equation of the hyperbola with eccentricity is $\sqrt{2}$ and the distance between whose foci is $16$ is $\frac{x^2}{k^2}-\frac{y^2}{32}=1$, then find the value of $k^2$.

Find the equation of the hyperbola whose vertices are $\left(0,\pm 3\right)$ and the eccentricity is $\frac{4}{3}$. Also find the coordinates of its foci.

The equation of the hyperbola whose foci are $\left(0,\pm 13\right)$ and the length of whose conjugate axis is $24$ is $\frac{y^2}{25}-\frac{x^2}{k^2}=1$, then find the value of $k$.

The equation of the hyperbola whose foci are $\left(0,\pm 10\right)$ and the length of whose latus rectum is $9 \mathrm{units}$ is $\frac{y^2}{k^2}-\frac{x^2}{36}=1$, then find the value of $k$.

The equation of the hyperbola having its foci at $\left(0,\pm \sqrt{14}\right)$ and passing through the point $\left(3,4\right)$ is $\frac{y^2}{k^2}-\frac{x^2}{7}=1$, then find the value of $k^2$.