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प्रश्न
Find the equation of the parabola in the cases given below:
Vertex (1, – 2) and Focus (4, – 2)
उत्तर
In given data the parabola is open rightwards and symmetric about the line parallel to x-axis.
Equation of parabola
(y – k)2 = 4a(x – h)
Vertex (h, k) = (1, – 2)
(y + 2)2 = 4a(x – 1)
a = AS = 3
Equation of parabola
(y + 2)2 = 4(3)(x – 1)
(y + 2)2 = 12(x – 1)
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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.
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Find the co-ordinates of the focus, vertex, equation of the directrix, axis and the length of latus rectum of the parabola
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Find the equation of the parabola which is symmetrical about x-axis and passing through (–2, –3).
The focus of the parabola x2 = 16y is:
The eccentricity of the parabola is:
The double ordinate passing through the focus is:
Find the equation of the ellipse in the cases given below:
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Find the equation of the hyperbola in the cases given below:
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y2 – 4y – 8x + 12 = 0
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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
