Find the equation of the ellipse in the cases given below: Foci e(±3,0),e+12 - Mathematics

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

Foci `(+- 3, 0), "e"+ 1/2`

बेरीज

Foci (± c, 0) = (± 3, 0)

e = `1/2`

c = ae = 3

`"a"(1/2)` = 3

a = 6 ⇒ a² = 36

b² = a² – c²

b² = 36 – 9 = 27

b² = 27

Equation of the ellipse be `x^2/"a"^2 + y^2/"b"^2` = 1

`x^2/36 + y^2/27` = 1

shaalaa.com

Conics

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

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

पाठ 5 Two Dimensional Analytical Geometry-II
Exercise 5.2 | Q 2. (i) | पृष्ठ १९६

The parabola y² = kx passes through the point (4, -2). Find its latus rectum and focus.

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

x² = - 16y

The focus of the parabola x² = 16y is:

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

Passes through (2, – 3) and symmetric about y-axis

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

End points of latus rectum (4, – 8) and (4, 8)

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

Foci (0, ±4) and end points of major axis are (0, ±5)

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

Centre (2, 1), one of the foci (8, 1) and corresponding directrix x = 4