Question

How do we find the latus rectum of the parabola `y=2x^2`?

Answer 1

Latus rectum is `1/2`.

Explanation:

Vertex form of equation of paarbola is `y=a(x-h)^2+k`, where vertex is `(0,0)` and `x=h` is the axis of symmetry. In the equation `y=a(x-h)^2+k`, focus is `(h,k+1/(4a))` and directrix is `y=k-1/(4a)`.

As such for `y=2x^2=2(x-0)^2+0`, while vertex is `(0,0)`

focus are `(0,0+1/8)` or `(0,1/8)` andpoints on latus rectum would be on bothsides of parabola from focus.

As `y=1/8` gives `x=+-1/4`, which means points of the latus rectum are `(-1/4,1/8)` and `(1/4,1/8)`

and latus rectum is `1/2`.

graph{(y-2x^2)(x^2+(y-1/8)^2-0.0001)((x+1/4)^2+(y-1/8)^2-0.0001)((x-1/4)^2+(y-1/8)^2-0.0001)=0 [-0.628, 0.622, -0.0875, 0.5375]}