Question

The lactus rectum of a parabola is a line segment passing through the focus of the parabola......?

The lactus rectum of a parabola is a line segment passing through the focus of the parabola such that the segment is parallel to the directrix and has both endpoints on the parabola. Show that the lactus rectum of the graph of the equation y=a(x-h)^2 +k has the length 1/a.

Answer 1

Pleasesee below.

Explanation:

The equation of parabola is `y=a(x-h)^2+k`

hence vertex is `(h,k)` and axis of symmetry is `x-h=0`

This is the equation of a vertical parabola. In such parabolas, for the equation is of the type `(x-h)^2=4p(y-k)`, the focus is `(h,k+p)` and the directrix is `y=h-p`.

So let us convert the given equation `y=a(x-h)^2+k` to this form and this becomes `(x-h)^2=1/a(y-k)`

Hence, focus is `(h,k+1/(4a))`.

As axis of symmetry is `x-h=0`, the line perpendicular to it (i.e. parallel to directrx) passing through focus `(h,k+1/(4a))` is `y=k+1/(4a)`

Let us find the two points, where this line `y=k+1/(4a)` intersects parabola `y=a(x-h)^2+k`. These will be given by

`a(x-h)^2+k=k+1/(4a)` or `(x-h)^2=1/(4a^2)`

i.e. `x-h=+-1/(2a)` and `x=h+1/(2a)` and `x=h-1/(2a)`

the two points at which line parallel to directrix cuts parabola are

`(h+1/(2a),k+1/(4a))` and `(h-1/(2a),k+1/(4a))`

As ordinate is same, length of latus rectum is

`h+1/(2a)-(h-1/(2a))`

= `h+1/(2a)-h+1/(2a)`

= `1/a`