Question

What is the length of latus rectum of a parabola, whose equation is `y^2+12x=0`?

Answer 1

Length of latus rectum is `12` units.

Explanation:

For a vertical parabola equation is `(y-k)=4a(x-h)^2` and for horizontal parabola it is `(x-h)=4a(y-k)^2` . The vertex (in both cases) is `(h,k)`.

In the equation `(y-k)=a(x-h)^2`, focus is `(h,k+1/(4a))` and directrix is `y=k-1/(4a)` and in case equation is `(x-h)=a(y-k)^2`, focus is `(h+1/(4a),k)`, directrix is `x=h-1/(4a)`.

Length of latus rectum is four times than the distance of focus from vertex.

Here equation is `y^2+12x=0` or `(x-0)=-1/12(y-0)^2` and hence `h=0`, `k=0` and `a=-1/12` or `1/(4a)=-3`

Hence vertex is `(0,0)` and focus is `(-3,0)`. As distance between vertex and focus is `3` units, length of latus rectum is `12` units.

graph{(y^2+12x)(x^2+y^2-0.1)((x+3)^2+y^2-0.1)((x+3)^2+(y-6)^2-0.1)((x+3)^2+(y+6)^2-0.1)=0 [-18, 14, -8, 8]}