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4. (02.08) derive the equation of the parabola with a focus at (6, 2) a…

Question

  1. (02.08) derive the equation of the parabola with a focus at (6, 2) and a directrix of y = 1. (2 points)

f(x)=-\frac{1}{2}(x - 6)^2+\frac{3}{2}
f(x)=\frac{1}{2}(x - 6)^2+\frac{3}{2}
f(x)=-\frac{1}{2}(x+\frac{3}{2})^2+6
f(x)=\frac{1}{2}(x+\frac{3}{2})^2+6

Explanation:

Step1: Recall the definition of a parabola

The distance from any point $(x,y)$ on the parabola to the focus $(6,2)$ is equal to the distance from the point $(x,y)$ to the directrix $y = 1$. The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, and the distance from the point $(x,y)$ to the line $y = k$ is $|y - k|$. So, $\sqrt{(x - 6)^2+(y - 2)^2}=|y - 1|$.

Step2: Square both sides

$(x - 6)^2+(y - 2)^2=(y - 1)^2$. Expand the equations: $(x - 6)^2+y^{2}-4y + 4=y^{2}-2y+1$.

Step3: Simplify the equation

Cancel out $y^{2}$ on both sides: $(x - 6)^2-4y + 4=-2y+1$. Move the terms with $y$ to one - side: $(x - 6)^2+4 - 1=-2y + 4y$. So, $(x - 6)^2+3 = 2y$.

Step4: Solve for $y$ (which is $f(x)$)

$y=\frac{1}{2}(x - 6)^2+\frac{3}{2}$, so $f(x)=\frac{1}{2}(x - 6)^2+\frac{3}{2}$.

Answer:

$f(x)=\frac{1}{2}(x - 6)^2+\frac{3}{2}$ (the second option)