हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा १२

Find the equation of the parabola in the cases given below: Vertex (1, – 2) and Focus (4, – 2) - Mathematics

Advertisements
Advertisements

प्रश्न

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

Vertex (1, – 2) and Focus (4, – 2)

योग
Advertisements

उत्तर


In given data the parabola is open rightwards and symmetric about the line parallel to x-axis.

Equation of parabola

(y – k)2 = 4a(x – h)

Vertex (h, k) = (1, – 2)

(y + 2)2 = 4a(x – 1)

a = AS = 3

Equation of parabola

(y + 2)2 = 4(3)(x – 1)

(y + 2)2 = 12(x – 1)

shaalaa.com
Conics
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Two Dimensional Analytical Geometry-II - Exercise 5.2 [पृष्ठ १९६]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 5 Two Dimensional Analytical Geometry-II
Exercise 5.2 | Q 1. (iii) | पृष्ठ १९६

संबंधित प्रश्न

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


The average variable cost of the monthly output of x tonnes of a firm producing a valuable metal is ₹ `1/5`x2 – 6x + 100. Show that the average variable cost curve is a parabola. Also, find the output and the average cost at the vertex of the parabola.


Find the axis, vertex, focus, equation of directrix and the length of latus rectum of the parabola (y - 2)2 = 4(x - 1)


The distance between directrix and focus of a parabola y2 = 4ax is:


The equation of directrix of the parabola y2 = -x 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 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:

Foci (± 2, 0), Eccentricity = `3/2`


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


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 = – 8x


Identify the type of conic and find centre, foci, vertices, and directrices of the following:

`y^2/16 - x^2/9` = 1


Prove that the length of the latus rectum of the hyperbola `x^2/"a"^2 - y^2/"b"^2` = 1 is `(2"b"^2)/"a"`


Identify the type of conic and find centre, foci, vertices, and directrices of the following:

`(x + 1)^2/100 + (y - 2)^2/64` = 1


Identify the type of conic and find centre, foci, vertices, and directrices of the following:

`(y - 2)^3/25 + (x + 1)^2/16` = 1


Choose the correct alternative:

The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half the distance between the foci is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×