Advertisements
Advertisements
प्रश्न
Find the vertex, focus, axis, directrix, and the length of the latus rectum of the parabola y2 – 8y – 8x + 24 = 0.
Advertisements
उत्तर
y2 – 8y – 8x + 24 = 0
⇒ y2 – 8y – 42 = 8x – 24 + 42
⇒ (y – 4)2 = 8x – 8
⇒ (y – 4)2 = 8(x – 1)
⇒ (y – 4)2 = 4(2) (x – 1)
∴ a = 2
Y2 = 4(2)X where X = x – 1 and Y = y – 4
| X, Y coordinates | x, y coordinates | |||
| Vertex (0, 0) | X = 0 | Y = 0 | x - 1 = 0 x = 1 |
y - 4 = 0 (1, 4) y = 4 |
| Focus (a, 0) | X = 2 | Y = 0 | x - 1 = 2 x = 2 + 1 = 3 |
y - 4 = 0 (3,4) y = 4 |
| Axis x-axis |
Y = 0 | y - 4 = 0 | y = 4 | |
| Directrix x + a = 0 |
X + 2 = 0 | x - 1 + 2 = 0 x + 1 = 0 |
x = - 1 | |
| Length of Latus rectum | 4a = 8 | |||
APPEARS IN
संबंधित प्रश्न
Find the co-ordinates of the focus, vertex, equation of the directrix, axis and the length of latus rectum of the parabola
y2 = 20x
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 axis, vertex, focus, equation of directrix and the length of latus rectum of the parabola (y - 2)2 = 4(x - 1)
The equation of directrix of the parabola y2 = -x is:
Find the equation of the ellipse in the cases given below:
Foci (0, ±4) and end points of major axis are (0, ±5)
Identify the type of conic and find centre, foci, vertices, and directrices of the following:
`y^2/16 - x^2/9` = 1
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:
18x2 + 12y2 – 144x + 48y + 120 = 0
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
The latus-rectum of a conic section is:
