Question

What is the equation for the parabola with the given focus and directrix?

Focus:(0,1/12). Directrix: y=-1/12.​

Answer 1

`x^2&=(1)/(3)y`

Explanation:

Given values:

`\bullet\;\;\textsf{Focus:}\quad \left(0, (1)/(12)\right)`

`\bullet\;\;\textsf{Directrix:} \quad y=-(1)/(12)`

The directrix of a parabola is a fixed line outside of the parabola that is perpendicular to the axis of symmetry. Therefore, as the directrix is horizontal (y = -1/12), the parabola has a vertical axis of symmetry. This means that the parabola is vertical (opens upwards or downwards).

The focus of a parabola is a fixed point located on the axis of symmetry, such that all points on the parabola are equidistant to both the focus and the directrix. It is located inside the parabola. Therefore, as the focus is above the directrix, the parabola opens upwards. The axis of symmetry of the given parabola is the x-value of the focus, x = 0.

The vertex of a parabola is its maximum or minimum point (turning point). As the parabola opens upwards, its vertex is a minimum. The vertex is located on the axis of symmetry. Therefore, the x-coordinate of the vertex of the given parabola is x = 0.

The standard form of a parabola with a vertical axis of symmetry is:

`\boxed{(x-h)^2=4p(y-k)}`

where:

• p ≠ 0
• Vertex = (h, k)
• Focus = (h, k+p)
• Directrix: y = (k - p)
• Axis of symmetry: x = h

As the focus is (0, 1/12):

`k+p=(1)/(12)`

As the directrix is y = -1/12:

`k-p=-(1)/(12)`

Add the two equations together to eliminate p, and solve for k:

`\begin{array}{crcccr}\vphantom{\frac12}&(k&+&p&=&(1)/(12)\\+&(k&-&p&=&-(1)/(12)\vphantom{\frac12}\\\cline{2-6}\vphantom{\frac12}&2k&&&=&0\end{array}`

Therefore, k = 0.

Substitute the found value of k into one of the equations and solve for p:

`0+p=(1)/(12)\implies p=(1)/(12)`

Substitute the values of h, k and p into the standard formula:

`\begin{aligned}(x-0)^2&=4\cdot (1)/(12)(y-0)\\\\x^2&=(1)/(3)y\end{aligned}`

Therefore, the equation of the parabola in standard form with the given focus and directrix is:

`\large\boxed{\boxed{x^2&=(1)/(3)y}}`

This can be rewritten in the standard form of a quadratic equation, y = ax² + bx + c, by multiplying both sides by 3 to give y = 3x².