English

Find the equation of the hyperbola with vertices (± 5, 0), foci (± 7, 0) - Mathematics

Advertisements
Advertisements

Question

Find the equation of the hyperbola with vertices (± 5, 0), foci (± 7, 0)

Sum

Solution

Given that vertices (± 5, 0), foci (± 7, 0)

Vertex of hyperbola = (± a, 0) and foci (± ae, 0)

∴ a = 5 and ae = 7

⇒ 5 × e = 7

⇒ e = `7/5`

Now b2 = a2(e2 – 1)

⇒ b2 = `25(49/25 - 1)`

⇒ b2 = `25 xx 24/25`

⇒ b2 = 24

The equation of the hyperbola is `x^2/25 - y^2/24` = 1

shaalaa.com
  Is there an error in this question or solution?
Chapter 11: Conic Sections - Exercise [Page 204]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 11
Chapter 11 Conic Sections
Exercise | Q 32.(a) | Page 204

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Find the equation of the hyperbola satisfying the given conditions:

Vertices (0, ±5), foci (0, ±8)


Find the equation of the hyperbola satisfying the given conditions:

Foci (0, ±13), the conjugate axis is of length 24.


Find the equation of the hyperbola satisfying the given conditions:

Foci `(+-3sqrt5, 0)`, the latus rectum is of length 8.


Find the equation of the hyperbola whose focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2 .


Find the equation of the hyperbola whose focus is (1, 1) directrix is 2x + y = 1 and eccentricity = \[\sqrt{3}\].


Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .

9x2 − 16y2 = 144


Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .

16x2 − 9y2 = −144


Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .

 4x2 − 3y2 = 36


Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .

 3x2 − y2 = 4 


Find the equation of the hyperbola whose  foci are (4, 2) and (8, 2) and eccentricity is 2.


Find the equation of the hyperbola whose vertices are at (± 6, 0) and one of the directrices is x = 4.


Find the equation of the hyperbola whose foci at (± 2, 0) and eccentricity is 3/2. 


Find the equation of the hyperbola satisfying the given condition:

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


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 length of the transverse axis along x-axis with centre at origin of a hyperbola is 7 and it passes through the point (5, –2). The equation of the hyperbola is ______.


If the distance between the foci of a hyperbola is 16 and its eccentricity is `sqrt(2)`, then obtain the equation of the hyperbola.


Find the equation of the hyperbola with vertices (0, ± 7), e = `4/3`


The distance between the foci of a hyperbola is 16 and its eccentricity is `sqrt(2)`. Its equation is ______.


Equation of the hyperbola with eccentricty `3/2` and foci at (± 2, 0) is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×