**$f\left(x\right)=-\dfrac{1}{16} \left(x\ -2\right)^2-2$****$f\left(x\right)=\ \dfrac{1}{16} \left(x\ -2\right)^2+2$****$f\left(x\right)=\ \dfrac{1}{16} \left(x\ -2\right)^2-2$****$f\left(x\right)=\ \dfrac{1}{16} {- \left(x\ +2\right)}^2-2$**

The aim of the question is to find the **quadratic function** of the given equations for which **directrix** and **focus** are given.

The basic concept behind this question is the knowledge of **parabola** and its equations as well as the **distance formula** between two points. The** distance formula** can be written as following for $2$ points $A= (x_1\ ,y_1)$ and $B = (x_2\ ,y_2)$

\[D_{AB}\ =\ \sqrt{\left(x_2-\ x_1\right)^2+\left(y_2-\ y_1\right)^2}\]

## Expert Answer

Given data we have:

**Directrix** $y = -2$

**Focus** $= (2, 6)$

Let us suppose a point $P = (x_1\ ,y_1)$ on the **parabola**.

And another point $Q = (x_2\ ,y_2)$ near the **directrix** of the **parabola**.

Using **distance formula** to find the distance between these two points $PQ$ and putting the **value of focus** in its equation, we get:

\[D_{PQ}\ =\ \sqrt{\left(x_2-\ x_1\right)^2+\left(y_2-\ y_1\right)^2}\]

Putting values in the above formula we get:

\[D_{PQ}\ =\ \sqrt{\left(x\ -2\right)^2+\left(y\ -6\right)^2}\]

As we know that in a **parabola**, all the points on it have **equal distance from the directrix** and as well as** focus**, so we can write for the value of the **directrix** as follows and put it equal to the **distance formula****:**

\[= y_2-\ y_1\]

\[=y-(-2) \]

Now putting equal to **distance formula**:

\[\sqrt{\left(x\ -2\right)^2+\left(y\ -6\right)^2}\ =\ \left|y-(-2)\ \right|\]

\[\sqrt{\left(x\ -2\right)^2+\left(y\ -6\right)^2}=\ \left|y+2\ \right|\]

Taking **square** on both the sides of equation:

\[\left(\sqrt{\left(x\ -2\right)^2+\left(y\ -6\right)^2}\right)^2=\left(\left|y+2\ \right|\right)^2\]

Solving the equations:

\[\left(x\ -2\right)^2+\left(y\ -6\right)^2\ =\ \left(y\ +\ 2\right)^2\]

\[\left(x\ -2\right)^2\ =\ \left(y\ +\ 2\right)^2-{\ \left(y\ -6\right)}^2\]

\[\left(x\ -2\right)^2\ =\ y^2+4y\ +4\ -y^2\ -36\ +12y\]

Cancelling out $y^2$:

\[\left(x\ -2\right)^2\ =\ 4y\ +12y\ +4\ -36\ \]

\[\left(x\ -2\right)^2\ =\ 16y\ +4\ -36\ \]

\[\left(x\ -2\right)^2\ =\ 16y\ -32\]

\[\left(x\ -2\right)^2+32\ =\ 16y\ \]

\[{\ 16y\ =\left(x\ -2\right)}^2+32\]

\[y\ =\frac{\left(x\ -2\right)^2}{16}+\frac{32}{16}\]

\[y\ =\frac{\left(x\ -2\right)^2}{16}+2\]

The required **quadratic equation** is:

\[ y\ =\frac{1}{16}\left(x\ -2\right)^2+2\ \]

## Numerical Results

By using the **directrix value** of $y = -2$ and **focus** of $(2,6)$ following **quadratic equation** is created:

\[y\ =\frac{1}{16}\left(x\ -2\right)^2+2\]

So from the $4$ given options, **option $2$ is correct**.

## Example

Using $y = -1$ as the **directrix value** and **focus** $(2,6)$ what will be the required **quadratic function**?

**Solution**:

**Directrix** $y = -1$

**Focus** $= (2, 6)$

Point $P = (x_1\ ,y_1)$ on the **parabola**.

Point $Q = (x_2\ ,y_2)$ near the **directrix** of the** parabola**.

Using **distance formula** to find the distance between these two points $PQ$ and putting the **value of focus** in its equation, we get:

\[D_{PQ}=\sqrt{\left(x-2\right)^2+\left(y-6\right)^2}\]

Value of **directrix** is:

\[= y_2-\ y_1\]

\[=y-(-1) \]

Now putting equal to **distance formula**:

\[\sqrt{\left(x\ -2\right)^2+\left(y\ -6\right)^2}=\ \left|y+1\ \right|\]

Taking square on both the sides:

\[\left(\sqrt{\left(x\ -2\right)^2+\left(y\ -6\right)^2}\right)^2=\left(\left|y+1\ \right|\right)^2\]

\[\left(x\ -2\right)^2+\left(y\ -6\right)^2\ =\ \left(y\ +\ 1\right)^2\]

\[\left(x-2\right)^2\ =\ \left(y\ +\ 1\right)^2-{\ \left(y\ -6\right)}^2\]

\[\left(x-2\right)^2\ =\ y^2+2y\ +1\ -y^2\ -36\ +12y\]

\[\left(x-2\right)^2\ =\ 2y\ +12y\ +1\ -36\ \]

\[\left(x-2\right)^2\ =\ 14y\ -35\]

\[{\ 14y=\left(x\ -2\right)}^2+35\]

\[y\ =\frac{\left(x\ -2\right)^2}{14}+\frac{35}{14}\]

\[y\ =\frac{1}{14} [\left(x\ -2\right)^2+35]\]

The required **quadratic equation** is:

\[y\ =\frac{1}{14} [\left(x\ -2\right)^2+35]\]