Question

determine the equation in standard form of the parabola that satisfies the given conditions
directrix at ( y = - 1 ); focus at ( ( - 5, - 3 ) )
question help: video ebook

Explanation:

Step1: Recall the definition of parabola

The distance from any point \((x,y)\) on the parabola to the focus \((- 5,-3)\) 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 a point \((x,y)\) to the line \(y = k\) is \(|y - k|\).
So, \(\sqrt{(x + 5)^2+(y + 3)^2}=|y+1|\).

Step2: Square both sides of the equation

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Another way:
The standard form of a parabola with a vertical directrix is \((x - h)^2 = 4p(y - k)\).
The mid - point between the focus \((-5,-3)\) and the directrix \(y=-1\) is the vertex \((h,k)\). The \(x\) - coordinate of the vertex \(h=-5\), and the \(y\) - coordinate of the vertex \(k=\frac{-3+( - 1)}{2}=-2\).
The distance between the focus and the vertex \(p=-3+2=-1\) (since the focus is below the vertex).
Substitute \(h=-5,k = - 2,p=-1\) into \((x - h)^2=4p(y - k)\)

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Answer:

\((x + 5)^2=-4(y + 2)\)