Find the equation of the parabola in the cases given below: End points of latus rectum (4, – 8) and (4, 8)

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

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

Sum

Focus = (4, 0)

Equation of the parabola will be of the form y² = 4ax

Here a = 4

⇒ y² = 16x

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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. (iv) | Page 196

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

y² = 20x

Find the axis, vertex, focus, equation of directrix and the length of latus rectum of the parabola (y - 2)² = 4(x - 1)

The eccentricity of the parabola 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:

Vertex (1, – 2) and Focus (4, – 2)

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:

y² – 4y – 8x + 12 = 0