Find an equation of the parabola with focus at (-4,0) and with directrix x=3.

Question

Express your answer in the form x=f(y)

So, I got (y-0)^2=4(3.5)(x-(-1/2))

I simplified it to be y^2=14(x+1/2)

and I solved for x which gave me x=1/14y^2-1/2

still says its wrong! :((

Answer 1

I think you got it all right, except for the sign of the coefficient of the y squared term. This parabola opens to the left and is in the 2nd and 3rd quadrants. So the sign has to be negative. It's -1/14, not +1/14

A parabola is the collection of all points (x,y) equidistant from a point (the focus) and a line (the directrix).

Let (x,y) be a point on the parabola, any point.

The distance from (x,y) to x=3 is the same as the distance from (x,y) to (-4,0)

Using the definition of a parabola, find an expression for the distance from (x,y) to the direction line

and to the focus point and set them equal

the distance formula for two points is d =( (x-x₁₎² + (y-y₁)²)^1/2

substituting for the focus = square root of (x+4)²+y² set this equal to the distance from (x,y) to the directrix line or square root of (x+4)² + y² = 3-x

square both sides to get (x+4)² + y² = 9-6x+x²

expand left side and cancel x² terms

x² + 8x + 16 + y² = x² - 6x + 9

14x=-y²-7 divide by 14 x=-y²/14-1/2 the coefficient of the y² term is -1/14 not +1/14

Other ways to work this, but you got a sign wrong somewhere

Answer 2

Hi Hira

If you're going to use (y-k)²=4p(x-k)², then make sure the sign of your p is correct. If the parabola opens to the left (as this one does), then the focus is to the left of the vertex, so p should be negative. Everything else seems on the right track.

You'll know this is the case if you draw it. But if you ever doubt the formula you're using as a shortcut, you can always go back to definitions! Since a parabola is the locus of all points the same distance from a focus and directrix, we can start there.

The distance from (x₀,y₀) to the directrix is |x₀-3|

The distance from (x₀,y₀) to the focus is sqrt((x₀-(-4))²+(y₀-0)²)

If these two distances are equal, then |x₀-3|=sqrt((x₀-(-4))²+(y₀-0)²)

(x-3)²=(x+4)²+(y)²

x²-6x+9=x²+8x+16+y²

-14x=y²+7

x=(-1/14)y²-1/2

Let me know if you need any further explanation!