Question

Find the equation of the parabola with its focus at (–5,0) and its directrix y = 2.Question options:A) y = –4(x + 5)^2 + 1B) y = –1∕4 (x + 5)^2 + 1C) y = –1∕4(x + 1)^2 + 5D) y = 1∕4(x + 5)^2 + 1

Answer 1

Recall that a parabola is a curve where any point is at an equal distance from a fixed point (the focus) and a fixed straight line (the directrix).

Let (x,y) a point on the parabola with its focus at (–5,0) and its directrix y = 2, then we can set the following equation:

`\sqrt[]{(x-(-5))^2+(y-0)^2}=2-y\text{.}`

Simplifying the above result we get:

`\sqrt[]{(x+5)^2+y^2}=2-y\text{.}`

Taking the above result to the power of 2 we get:

`\begin{gathered} \sqrt[]{(x+5)^2+y^2}^2=(2-y)^2, \\ (x+5)^2+y^2=(2-y)^2\text{.} \end{gathered}`

Therefore:

`(x+5)^2+y^2=4-4y+y^2\text{.}`

Subtracting y² from the above equation we get:

`\begin{gathered} (x+5)^2+y^2-y^2=4-4y+y^2\text{-}y^2, \\ (x+5)^2=4-4y^{}\text{.} \end{gathered}`

Subtracting 4 from the above equation we get:

`\begin{gathered} (x+5)^2-4=4-4y-4, \\ (x+5)^2-4=-4y\text{.} \end{gathered}`

Finally, dividing the above equation by -4 we get:

`\begin{gathered} ((x+5)^2-4)/(-4)=(-4y)/(-4), \\ y=-(1)/(4)(x+5)^2+1. \end{gathered}`

Answer: Option B.