Question

Question #6e900

Answer 1

The equation is: `x^2- 2xy+y^2+2x-6y+3 = 0`

Explanation:

We need the formula for the Distance from a Point to a Line :

d = `|ax+by+c|/sqrt(a^2+b^2)`

where the point is `(x,y)` and the equation of the line is of the form `ax+by+c=0`.

We are given the line `x+y+1 = 0`, therefore, the distance from any point `(x,y)` on the parabola to the line is:

`d = |x+y+1|/sqrt(1^2+1^2)`

`d = |x+y+1|/sqrt2" [1]"`

The distance from the focus `(-1,1)` to any point `(x,y)` on the parabola is:

`d = sqrt((x+1)^2+(y-1)^2)" [2]"`

Because a parabola is defined at the locus of points equidistant from and point and a line, we can set the right side of equation [1] equal to the right side of equation [2]:

`|x+y+1|/sqrt2 = sqrt((x+1)^2+(y-1)^2)`

Square both sides:

`1/2(x+y+1)^2 = (x+1)^2+(y-1)^2`

Expand the squares:

`1/2(x^2+2xy+y^2+2x+2y+1) = x^2+2x+1 + y^2-2y+1`

Use the distributive property:

`1/2x^2+xy+1/2y^2+x+y+1/2 = x^2+2x+1 + y^2-2y+1`

To write in the General Cartesian Form for a Conic Section , `Ax^2+Bxy+Cy^2+Dx+Ey + F = 0`, we combine like terms::

`1/2x^2-xy+1/2y^2+x-3y+3/2= 0`

Multiply both sides by 2:

`x^2- 2xy+y^2+2x-6y+3 = 0`

Here is a graph of, the parabola, the focus, and the directrix: