 # Using a directrix of y=−2 and a focus of (2, 6), what quadratic function is created? 1.  $f\left(x\right)=-\dfrac{1}{16} \left(x\ -2\right)^2-2$
2.  $f\left(x\right)=\ \dfrac{1}{16} \left(x\ -2\right)^2+2$
3.  $f\left(x\right)=\ \dfrac{1}{16} \left(x\ -2\right)^2-2$
4.  $f\left(x\right)=\ \dfrac{1}{16} {- \left(x\ +2\right)}^2-2$

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 $B = (x_2\ ,y_2)$

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

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\$

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

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

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

$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\$

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

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

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

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

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