Find the equation of the hyperbola in the cases given below: Passing through (5, – 2) and length of the transverse axis along x-axis and of length 8 units - Mathematics

Find the equation of the hyperbola in the cases given below:

Passing through (5, – 2) and length of the transverse axis along x-axis and of length 8 units

योग

Transverse axis along x-axis

`x%2/"a"^2 - y^2/"b"^2` = 1

Length of transverse axis 2a = 8

⇒ a = 4

`x^2/16 - y^2/"b"^2` = 1

At (5, – 2) `25/16 - 4/"b"^2` = 1

`25/16 - 1 = 4/"b"^2`

`(25 - 16)/16 = 4/"b"^2`

⇒ `9/16 = 4/"b"^2`

b² = `(16 xx 4)/9` = 4

Equation of hyperbola `x^2/16 - y^2/(64/9)` = 1

`x%^2/16 - (9y^2)/64` = 1

shaalaa.com

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

अध्याय 5: Two Dimensional Analytical Geometry-II - Exercise 5.2 [पृष्ठ १९६]

अध्याय 5 Two Dimensional Analytical Geometry-II
Exercise 5.2 | Q 3. (iii) | पृष्ठ १९६

Find the equation of the parabola whose focus is the point F(-1, -2) and the directrix is the line 4x – 3y + 2 = 0.

Find the vertex, focus, axis, directrix, and the length of the latus rectum of the parabola y² – 8y – 8x + 24 = 0.

Find the co-ordinates of the focus, vertex, equation of the directrix, axis and the length of latus rectum of the parabola

x² = 8y

Find the co-ordinates of the focus, vertex, equation of the directrix, axis and the length of latus rectum of the parabola

x² = - 16y

Find the equation of the parabola which is symmetrical about x-axis and passing through (–2, –3).

Find the equation of the parabola in the cases given below:

Vertex (1, – 2) and Focus (4, – 2)

Find the equation of the hyperbola in the cases given below:

Foci (± 2, 0), Eccentricity = `3/2`

Find the vertex, focus, equation of directrix and length of the latus rectum of the following:

y² = 16x

Find the vertex, focus, equation of directrix and length of the latus rectum of the following:

y² – 4y – 8x + 12 = 0