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

Jul 28, 2017

Latus rectum is $\frac{1}{2}$.

#### Explanation:

Vertex form of equation of paarbola is $y = a {\left(x - h\right)}^{2} + k$, where vertex is $\left(0 , 0\right)$ and $x = h$ is the axis of symmetry. In the equation $y = a {\left(x - h\right)}^{2} + k$, focus is $\left(h , k + \frac{1}{4 a}\right)$ and directrix is $y = k - \frac{1}{4 a}$.

As such for $y = 2 {x}^{2} = 2 {\left(x - 0\right)}^{2} + 0$, while vertex is $\left(0 , 0\right)$

focus are $\left(0 , 0 + \frac{1}{8}\right)$ or $\left(0 , \frac{1}{8}\right)$ andpoints on latus rectum would be on bothsides of parabola from focus.

As $y = \frac{1}{8}$ gives $x = \pm \frac{1}{4}$, which means points of the latus rectum are $\left(- \frac{1}{4} , \frac{1}{8}\right)$ and $\left(\frac{1}{4} , \frac{1}{8}\right)$

and latus rectum is $\frac{1}{2}$.

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]}