What is the standard form of the equation of the parabola with a directrix at x=-9 and a focus at (8,4)?

Question

Genesis Allen · Accepted Answer

The equation of the parabola is #(y-4)^2=17(2x+1)#

Any point #(x,y)# on the parabola is equidistant from the directrix and the focus.

Therefore,

#x-(-9)=sqrt((x-(8))^2+(y-(4))^2)#

#x+9=sqrt((x-8)^2+(y-4)^2)#

Squaring and developing the #(x-8)^2# term and the LHS

#(x+9)^2=(x-8)^2+(y-4)^2#

#x^2+18x+81=x^2-16x+64+(y-4)^2#

#(y-4)^4=34x+17=17(2x+1)#

The equation of the parabola is #(y-4)^2=17(2x+1)#

graph{((y-4)^2-34x-17)((x-8)^2+(y-4)^2-0.05)(y-1000(x+9))=0 [-17.68, 4.83, -9.325, 1.925]}

Jace Anderson · Answer

undefined The standard form of the equation of a parabola with a horizontal axis of symmetry is given by:

$$ (x-h)^2 = 4p(y-k) $$

Where:
- (h, k) is the vertex of the parabola.
- The focus of the parabola is located at (h+p, k).
- The directrix of the parabola is given by the equation y = k - p.

Given that the focus is at (8,4) and the directrix is x = -9, we can determine:
- The vertex (h, k) as the midpoint between the focus and the directrix: (8+(-9))/2 = -0.5
- The distance between the vertex and the focus (p) is the absolute value of the difference between the x-coordinate of the focus and the x-coordinate of the vertex: |8 - (-0.5)| = 8.5

Therefore, the vertex is (-0.5, k) and the value of k is the y-coordinate of the focus, which is 4. So, the vertex is (-0.5, 4).

Plugging these values into the standard form equation, we get:

$$ (x + 0.5)^2 = 4 \cdot 8.5(y - 4) $$

$$ (x + 0.5)^2 = 34(y - 4) $$

This is the standard form of the equation of the parabola with a directrix at x = -9 and a focus at (8,4).