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)

Sum

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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Chapter 5: Two Dimensional Analytical Geometry-II - Exercise 5.2 [Page 196]

Chapter 5 Two Dimensional Analytical Geometry-II
Exercise 5.2 | Q 1. (iii) | Page 196

Find the equation of the parabola whose focus is the point F(-1, -2) and the directrix is the line 4x – 3y + 2 = 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

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 equation of directrix of the parabola y² = -x is:

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

Focus (4, 0) and directrix x = – 4

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:

x² = 24y

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

y² – 4y – 8x + 12 = 0