Question

suppose that the point $(-3,-3)$ is the focus and the line $x = -5$ is the directrix. derive the equation of the parabola. drag your answers to the correct boxes. distance from a point $(x,y)$ on the parabola to the focus: $d_1 =$ distance from $(x,y)$ to the directrix: $d_2 =$ write an equation using the distances to get rewrite the equation in vertex form as $y =$

Explanation:

Step1: 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, \((x_1,y_1)=(x,y)\) and \((x_2,y_2)=(-3,-3)\), so \(d_1=\sqrt{(x + 3)^2+(y + 3)^2}\).

Step2: Distance formula for \(d_2\)

The distance from a point \((x,y)\) to the vertical line \(x = a\) is \(|x - a|\). Here \(a=-5\), so \(d_2=|x+5|\).

Step3: Set \(d_1 = d_2\)

Since for a parabola \(d_1=d_2\), we have \(\sqrt{(x + 3)^2+(y + 3)^2}=|x + 5|\). Squaring both sides: \((x + 3)^2+(y + 3)^2=(x + 5)^2\).
Expand: \(x^{2}+6x + 9+(y + 3)^2=x^{2}+10x+25\).
Simplify: \((y + 3)^2=x^{2}+10x + 25-(x^{2}+6x + 9)\).
\((y + 3)^2=4x+16\).
Solve for \(x\): \(x=\frac{1}{4}(y + 3)^2-4\).

Answer:

\(d_1=\sqrt{(x + 3)^2+(y + 3)^2}\), \(d_2=|x + 5|\), \(\sqrt{(x + 3)^2+(y + 3)^2}=|x + 5|\), \(x=\frac{1}{4}(y + 3)^2-4\)