Question

Opens upward with focus 5 units away from the vertex.

Short answer

The equation of parabola having a vertex at the origin and opens upwards with the focus 5 units away from the vertex -$x^2=20y\text{.}$

Step-by-step solution

Step 1.Given Information.

The parabola vertex is at the origin and it opens upwards with a focus 5 units away from the vertex.

Step 2.The equation of a parabola that satisfies the conditions.

The parabola opens upwards so p is positive and focus is on positive y -axis, the coordinates of focus are obtained.

On substituting the value of p in the equation to obtain an equation$x^2=4ay$of a parabola. The parabola opens upwards so is positive and focus is on positive y-axis, the coordinates of focus are obtained an equation of parabola

The vertex at the origin i.e.,$(0,0)$

The parabola opens upwards so, p>0.

The focus is 5 away from the focus so, the coordinate of focus is(0,5)andp=5.

Now, substituting the value ofpin equation $x^2=4py$, we get the equation as –

$\begin{array}{l}{x}^2=4\left(5\right)y\\ \Rightarrow x^2=20y.\end{array}$