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.
Advertisements
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
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
y2 = 20x
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 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)
Find the vertex, focus, equation of directrix and length of the latus rectum of the following:
y2 = 16x
Identify the type of conic and find centre, foci, vertices, and directrices of the following:
`x^2/25 - y^2/144` = 1
The latus-rectum of a conic section is:
The fixed straight line used in the definition of a conic section is called the:
