Question

Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1.

Answer 1

Answer:

The equation of the parabola with a focus at (-5,5) and a directrix of y = -1 is $y = \frac{1}{12}\cdot (x+5)^{2}+2$.

Step-by-step explanation:

From statement we understand that parabola has its axis of symmetry in an axis parallel to y-axis. According to Analytical Geometry, the minimum distance between focus and directrix equals to twice the distance between vertex and any of endpoints.

If endpoints are (-5, 5) and (-5, -1), respectively, then such distance ($r$), dimensionless, is calculated by means of the Pythagorean Theorem:

$r = \frac{1}{2}\cdot \sqrt{[-5-(-5)]^{2}+[5-(-1)]^{2}}$

$r = 3$

And the location of the vertex ($V(x,y)$), dimensionless, which is below the focus, is:

$V(x,y) = F(x,y)-R(x,y)$ (1)

Where:

$F(x,y)$ - Focus, dimensionless.

$R(x,y)$ - Vector distance, dimensionless.

If we know that $F(x,y) = (-5,5)$ and $R(x,y) = (0,3)$, then the location of the vertex is:

$V(x,y) = (-5,5)-(0,3)$

$V(x,y) =(-5,2)$

In addition, we define a parabola by the following expression:

$y-k = \frac{(x-h)^{2}}{4\cdot r}$ (2)

Where:

$h$, $k$ - Coordinates of the vertex, dimensionless.

$r$ - Distance of the focus with respect to vertex, dimensionless.

If we know that $h = -5$, $k = 2$ and $r = 3$, then the equation of the parabola is:

$y = \frac{1}{12}\cdot (x+5)^{2}+2$

The equation of the parabola with a focus at (-5,5) and a directrix of y = -1 is $y = \frac{1}{12}\cdot (x+5)^{2}+2$.