Question

derive the equation of a parabola with a focus at (3,3) and a directrix at y = 1. enter your answers in the boxes. be sure to include the correct signs. y = □(x □)^2 □

Explanation:

Step1: Recall the definition of a parabola

A parabola is the set of all points $(x,y)$ that are equidistant from the focus $(x_f,y_f)=(3,3)$ and the directrix $y = 1$. The distance between the point $(x,y)$ and the focus is $d_1=\sqrt{(x - 3)^2+(y - 3)^2}$, and the distance between the point $(x,y)$ and the directrix $y = 1$ is $d_2=\vert y - 1\vert$. Since $d_1=d_2$, we have $\sqrt{(x - 3)^2+(y - 3)^2}=\vert y - 1\vert$.

Step2: Square both sides

Squaring both sides of the equation $\sqrt{(x - 3)^2+(y - 3)^2}=\vert y - 1\vert$ gives us $(x - 3)^2+(y - 3)^2=(y - 1)^2$.

Step3: Expand the squares

Expand the left - hand side: $(x - 3)^2+(y - 3)^2=x^{2}-6x + 9+y^{2}-6y+9$. Expand the right - hand side: $(y - 1)^2=y^{2}-2y + 1$. So, $x^{2}-6x + 9+y^{2}-6y+9=y^{2}-2y + 1$.

Step4: Simplify the equation

Cancel out $y^{2}$ on both sides: $x^{2}-6x + 9-6y+9=-2y + 1$. Move all terms involving $y$ to one side: $-6y + 2y=-x^{2}+6x-9 - 9+1$. Combine like terms: $-4y=-x^{2}+6x - 17$.

Step5: Solve for y

Divide both sides by $-4$: $y=\frac{1}{4}(x - 3)^2+2$.

Answer:

$y=\frac{1}{4}(x - 3)^2+2$