Find the equation of the parabola in the cases given below: Vertex (1, – 2) and Focus (4, – 2)

Find the equation of the parabola in the cases given below:

Vertex (1, – 2) and Focus (4, – 2)

योग

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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अध्याय 5: Two Dimensional Analytical Geometry-II - Exercise 5.2 [पृष्ठ १९६]

अध्याय 5 Two Dimensional Analytical Geometry-II
Exercise 5.2 | Q 1. (iii) | पृष्ठ १९६

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 = 8y

The focus of the parabola x² = 16y is:

The eccentricity of the parabola is:

The distance between directrix and focus of a parabola y² = 4ax is:

Find the equation of the parabola in the cases given below:

Passes through (2, – 3) and symmetric about y-axis

Find the equation of the parabola in the cases given below:

End points of latus rectum (4, – 8) and (4, 8)

Find the vertex, focus, equation of directrix and length of the latus rectum of the following:

x² = 24y

Find the vertex, focus, equation of directrix and length of the latus rectum of the following:

y² – 4y – 8x + 12 = 0