Advertisements
Advertisements
प्रश्न
The parabola y2 = kx passes through the point (4, -2). Find its latus rectum and focus.
उत्तर
y2 = kx passes through (4, -2)
(-2)2 = k(4)
⇒ 4 = 4k
⇒ k = 1
y2 = x = 4`(1/4)`x
a = `1/4`
Equation of LR is x = a or x – a = 0
i.e., x = `1/4`
⇒ 4x = 1
⇒ 4x – 1 = 0
Focus (a, 0) = `(1/4,0)`
APPEARS IN
संबंधित प्रश्न
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 co-ordinates of the focus, vertex, equation of the directrix, axis and the length of latus rectum of the parabola
x2 = 8y
The distance between directrix and focus of a parabola y2 = 4ax is:
Find the equation of the ellipse in the cases given below:
Length of latus rectum 4, distance between foci `4sqrt(2)`, centre (0, 0) and major axis as y-axis
Identify the type of conic and find centre, foci, vertices, and directrices of the following:
`x^2/25 + y^2/9` = 1
Identify the type of conic and find centre, foci, vertices, and directrices of the following:
`x^2/3 + y^2/10` = 1
Identify the type of conic and find centre, foci, vertices, and directrices of the following:
`x^2/25 - y^2/144` = 1
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 - 3)^2/225 + (y - 4)^2/289` = 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
