Question

Find the equation of the parabola with its focus at (6,2) and its directrix y = 0.

Answer 1

As per given by the question,

There are given that focus point at (6, 2) and line y=0.

Now,

Distance from line dL=distance from the point dF

Then,

$y=\sqrt[]{(x-x_1)^2+(y-y_1)^2^{}_{}}^{}$

Now,

$\begin{gathered} y=\sqrt[]{(x-x_1)^2+(y-y_1)^2^{}_{}}^{} \\ y=\sqrt[]{(x-6)^2+(y-2)^2} \end{gathered}$

Then,

Square on the both side of the equation,

$\begin{gathered} y^{}=\sqrt[]{(x-6)^2+(y-2)^2} \\ y^2=(x-6)^2+(y-2)^2 \end{gathered}$

According to the question,

There are given that y=0,

So,

$\begin{gathered} y^2=(x-6)^2+(y-2)^2 \\ y^2=x^2+36-12x+y^2+4-4y \\ y^2=x^2+y^2-12x-4y+40 \end{gathered}$

Now,

$\begin{gathered} x^2-12x-4y+40=0 \\ x^2-12x+40=4y \end{gathered}$

Now, find the value of jy from above equation;

$\begin{gathered} 4y=x^2-12x+40 \\ y=\frac{1}{4}(x^2-12x+40) \end{gathered}$

So,

$\begin{gathered} y=\frac{1}{4}((x-6)^2+4) \\ y=\frac{1}{4}(x-6)^2+1 \end{gathered}$

Hence, the option D is correct.