PUC Science कक्षा ११ - Karnataka Board PUC Question Bank Solutions

Find the equation of the hyperbola satisfying the given condition :

vertices (0, ± 3), foci (0, ± 5)

Find the equation of the hyperbola satisfying the given condition :

 foci (0, ± 13), conjugate axis = 24

find the equation of the hyperbola satisfying the given condition:

 vertices (± 7, 0), \[e = \frac{4}{3}\]

Find the equation of the hyperbola satisfying the given condition:

 foci (0, ± \[\sqrt{10}\], passing through (2, 3).

Show that the set of all points such that the difference of their distances from (4, 0) and (− 4,0) is always equal to 2 represents a hyperbola.

Write the equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0).

Equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0), is

The difference of the focal distances of any point on the hyperbola is equal to

The foci of the hyperbola 9x² − 16y² = 144 are

The equation of the hyperbola whose foci are (6, 4) and (−4, 4) and eccentricity 2, is

The foci of the hyperbola 2x² − 3y² = 5 are

The equation of the hyperbola whose centre is (6, 2) one focus is (4, 2) and of eccentricity 2 is

Find the axes, eccentricity, latus-rectum and the coordinates of the foci of the hyperbola 25x² − 36y² = 225.

Find the equation of the hyperbola satisfying the given condition :

 foci (± \[3\sqrt{5}\] 0), the latus-rectum = 8

find the equation of the hyperbola satisfying the given condition:

 foci (± \[3\sqrt{5}\]  0), the latus-rectum = 8

find the equation of the hyperbola satisfying the given condition:

foci (0, ± 12), latus-rectum = 36

Write the eccentricity of the hyperbola whose latus-rectum is half of its transverse axis.

If the latus-rectum through one focus of a hyperbola subtends a right angle at the farther vertex, then write the eccentricity of the hyperbola.