Can you help me come up with an equation for a parabola with vertex at (-2,4) and focus at (-4,1) ?

Question

I am stuck on this problem because I am not sure how to rotate the parabola. I used the equation x=a(y-k)^2+h with (h,k)=(-2,4). I think this problem has something to do with rotation but I'm not sure how to go about it. I appreciate the help!

Please help me come up with an equation. I asked this question before and I was told that the parabola was rotated by 52 degrees but I have no idea how to incorporate this info into the vertex formula.

Answer 1

I will give you some hints. There are several ways to find the equation. Perhaps the most informative is to use the definition of a parabola, i.e. for any point on the parabola, its distance to the focus is equal to its distance to the directrix.

To find the equation of the directrix:

Its slope will be the negative reciprocal of the slope of the axis of symmetry. The slope of the axis of symmetry it 3/2 so the slope of the directrix is -2/3. The distance from the vertex to the focus is the same as the distance from the vertex to the directrix. The distance from the vertex to the focus is √13. Find the point on the axis of symmetry that is √13 from the vertex (the equation of the axis of symmetry is y-1 = 3/2(x+4)). Also might have to use the distance formula or the equation of a circle with center (-2,4) and radius √13.

The key is finding the equation of the directrix. Once you have that there is a standard formula for the distance from a point to a line. If the equation of the line is ax+by+c = 0, then the distance to that line from some point P(x₁,y₁) is |ax₁+by₁+c|/√(a²+b²).

The distance from P to focus is √[(x₁+4)²+(y-1)²] that must equal the its distance to the directrix. Set those expressions equal, square both sides to remove radical, expand binomials, combine similar terms, setting equal to 0 and you will have your equation.

The general form for a conic section: Ax²+Bxy+Cy²+Dx+Ey+F = 0 is a parabola if A, B, C ≠ 0 and B²-4AC = 0.

Answer 2

I concur with Doug, but add some more to simplify.

the directix has equation,,,, -2/3=[y-7//[x+0]

the point (0,7) is twice as far from focus as vertex on same line so is on directix

y=-(2/3)x+7

m=-2/3,,,b=+7

the distance from point (x,y) to line is (if line above point vertical)

d_d=[mx+b-y]/√(1+m²)

the distance from point to focus is as Doug wrote,

d_f=√[(x₁+4)²+(y-1)²]

squaring both sides and plugging in values for m and b

(-2/3x+7-y)²/(1+4/9)=(x+4)²+(y-1)²

you need to check the algebra,

x²-(4/3)xy+(4/9)y²+(188/9)x+(100/9)y-(163/9)=0