Advertisements
Advertisements
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.
उत्तर
F(-1, -2)
l : 4x – 3y + 2 = 0
Let P(x, y) be any point on the parabola.
FP = PM
⇒ FP2 = PM2
⇒ (x + 1)2 + (y + 2)2 = `[(4x - 3y + 2)/(sqrt(4^2 + (-3)^2))]^2`
⇒ x2 + 2x + 1 + y2 + 4y + 4 = `(16x^2 + 9y^2 + 4 - 24xy + 16x - 12y)/(16 + 9)`
⇒ 25(x2 + y2 + 2x + 4y + 5) = 16x2 + 9y2 – 24xy + 16x – 12y + 4
⇒ (25 – 16)x2 + (25 – 9)y2 + 24xy + (50 – 16)x + (100 + 12)y + 125 – 4 = 0
⇒ 9x2 + 16y2 + 24xy + 34x + 112y + 121 = 0
APPEARS IN
संबंधित प्रश्न
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 = 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.
The profit ₹ y accumulated in thousand in x months is given by y = -x2 + 10x – 15. Find the best time to end the project.
The equation of directrix of the parabola y2 = -x is:
Find the equation of the parabola in the cases given below:
Focus (4, 0) and directrix x = – 4
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
