What is the equation of the parabola with a focus at (-3,4) and a directrix of y= -7?

Question

Eli Assouline · Accepted Answer

#y = 1/22(x- (-3))^2 -3/2#

We can determine that the vertex has the same x coordinate as the focus and a y coordinate that is the midpoint between the directrix and the focus because the directrix is a horizontal line:

#(h,k) = (-3,(4+ (-7))/2)#

#(h,k) = (-3,-3/2)#

Enter the values of h and k into the vertex form of the equation for a vertically opening parabola.

#y = a(x-h)^2+k#

#y = a(x- (-3))^2 -3/2#

Compute the signed vertical distance, #f#, from the vertex to the focus:

#f = 4 - (-3/2)#

#f = 11/2#

We know that #a = 1/(4f)#

#a = 1/(4(11/2))#

#a = 2/(4(11))#

#a = 1/22#

Enter the value in place of "a" in the equation:

#y = 1/22(x- (-3))^2 -3/2#

Eva Adamson · Answer

undefined The equation of a parabola with focus $(a, b + p)$ and directrix $y = b - p$ is given by:

$$ (x - a)^2 = 4p(y - b) $$

In this case, the focus is $(-3, 4)$ and the directrix is $y = -7$.

Comparing this to the general form, we can see that $a = -3$ and $b = 4$.

Since the directrix is $y = -7$, we can see that $b - p = -7$. So, $4 - p = -7$, which implies $p = 11$.

Substituting these values into the equation of the parabola, we get:

$$ (x + 3)^2 = 4 \cdot 11(y - 4) $$

$$ (x + 3)^2 = 44(y - 4) $$

Therefore, the equation of the parabola is $ (x + 3)^2 = 44(y - 4) $.