Find the equation of the parabola in the cases given below: Focus (4, 0) and directrix x = – 4

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

Focus (4, 0) and directrix x = – 4

योग

Focus (4, 0) and directrix x = – 4

Parabola is open rightwards vertex (0, 0)

a = 4

Distance AS = 4 unit

F² = 4(4)x

Equation of parabola

y² = 16x.

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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. (i) | पृष्ठ १९६

The parabola y² = kx passes through the point (4, -2). Find its latus rectum and focus.

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:

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

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 = – 8x

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

y² – 4y – 8x + 12 = 0