Advertisements
Advertisements
Question
Find the equation of the parabola in the cases given below:
Passes through (2, – 3) and symmetric about y-axis
Advertisements
Solution

x2 = 4ay
It passes through (2, – 3)
⇒ 22 = 4a(– 3)
4 = – 12a
⇒ a = `- 1/3`
⇒ 4a = `- 4/3`
∴ Equation of parabola is x2 = `- 4/3`y
3x2= – 4y
APPEARS IN
RELATED QUESTIONS
Find the equation of the parabola whose focus is the point F(-1, -2) and the directrix is the line 4x – 3y + 2 = 0.
The parabola y2 = kx passes through the point (4, -2). Find its latus rectum and focus.
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 = - 16y
The double ordinate passing through the focus is:
The equation of directrix of the parabola y2 = -x is:
Find the equation of the parabola in the cases given below:
Vertex (1, – 2) and Focus (4, – 2)
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 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
Find the equation of the hyperbola in the cases given below:
Foci (± 2, 0), Eccentricity = `3/2`
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:
`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
Show that the absolute value of difference of the focal distances of any point P on the hyperbola is the length of its transverse axis
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:
18x2 + 12y2 – 144x + 48y + 120 = 0
If the eccentricity e > 1, the conic section is:
