Question

derive the equation of a parabola with a focus at (3,2) and a directrix at y = -1. move one answer to each box. find the distance from a point (x,y) on the parabola to the focus d1 = distance from (x,y) to the directrix d2 = d1 = d2 →

Explanation:

Step1: Apply distance formula for $d_1$

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Here, the focus is $(3,2)$ and the point on the parabola is $(x,y)$, so $d_1=\sqrt{(x - 3)^2+(y - 2)^2}$.

Step2: Find distance for $d_2$

The directrix is $y=-1$. The distance from the point $(x,y)$ to the horizontal line $y = k$ is given by $|y - k|$. So, $d_2=|y+ 1|$.

Step3: Set $d_1 = d_2$ and simplify

Since $d_1 = d_2$, we have $\sqrt{(x - 3)^2+(y - 2)^2}=|y + 1|$. Squaring both sides to get rid of the square - root and the absolute - value (since squaring makes both sides non - negative), we obtain $(x - 3)^2+(y - 2)^2=(y + 1)^2$. Expand the equations: $(x - 3)^2+y^{2}-4y + 4=y^{2}+2y+1$. Then, $(x - 3)^2=y^{2}+2y + 1-(y^{2}-4y + 4)$. Simplify the right - hand side: $(x - 3)^2=y^{2}+2y + 1 - y^{2}+4y-4=(x - 3)^2=6y - 3$.

Answer:

$d_1=\sqrt{(x - 3)^2+(y - 2)^2}$, $d_2=|y + 1|$, $d_1 = d_2
ightarrow(x - 3)^2=6y - 3$