Please explain, I really need to learn

'Equidistant from a point and a line' tells you the curve is a parabola with its focus at the point. The line is its directrix. Since the line is vertical and the focus is to its left, the parabola opens to the left. It will have the general equation

.. x = (1/(4p))(y- k)^2 +h

where p is the distance from vertex to directrix and (h, k) is the vertex.

The vertex is half way between the focus (-3, 4) and the line x=1, so is

.. (h, k) = ((-3+1)/2, 4) = (-1, 4)

The distance from this to the directrix is p = (-1 -1) = -2, so the equation is

.. x = (-1/8)*(y -4)^2 -1

P is the locus of the point (x, y)

The equidistance from S(- 3, 4) (foci s(h, k ± a)

T the line containing (1, - 5) and (1, 5)

The general equation of parabola

(y - k)² = 4a(x - h)

P is the distance from vertex to directrix and (h, k) is the vertex

Foci (- 3, 4) and the line x = 1 (line x = 1(1, 0)

(h, k) = (x₁ + x₂/2, y₁ + y₂/2)

(h, k) = (- 3 + 1/2, 4 + 0/2)

(h, k) = (- 1, 2)

The distance from this to the directrix is p = ( - 1, - 1) = - 2

So equation (y - k)² = 4a(x - h)

(y - 2)² = 4(- 2) (x + 1)

(y - 2)² = - 8(x + 1)

Y² - 4y + 4 = - 8x - 8

Y² - 4y + 8x + 12 = 0

The equation will be for a parabola, because a parabola is equidistant to a line and a point not on that line. And that point is not on that line. Because that line is x=1.

Your textbook should explain how to formulate a parabola given the point (focus) and the line (directrix).

The line that contains points (1,-5) and (1,5) is the line x = 1

So we have a parabola with focus (−3,4) and directrix: x = 1

Vertex is halfway between focus and directrix ---> vertex = (−1,4)

p = Fx − Vx = −3 − (−1) = −2

Equation of parabola with horizontal axis of symmetry:

4p (x − h) = (y − k)², where (h, k) = vertex

−8 (x + 1) = (y − 4)²

In vertex form we get:

x = −1/8 (y − 4)² − 1

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Alternate method:

All points (x,y) that are equidistant from (−3,4) and the line x = 1:

Distance from (x,y) to (−3,4) = Distance from (x,y) to line x = 1

√((x+3)²+(y−4)²) = √((x−1)²)

(x+3)² + (y−4)² = (x−1)²

x² + 6x + 9 + (y−4)² = x² − 2x + 1

(y − 4)² = −8x − 8

(y − 4)² = −8 (x + 1)

In vertex form we get:

x = −1/8 (y − 4)² − 1

the graph/equation will yield a parabola with directrix x = 1.......required is if (x,y) on curve then ( x + 3 )² + ( y - 4 )² = ( x - 1 )² .......6 x + 9 + ( y - 4)² = - 2x + 1 ====> 8 x + 8 = - (y - 4)² =====> - 8 [ x + 1 ] = [y - 4 ]²..vertex at ( - 1 , 4 ) ; p = - 2 , thus open "parallel " to the negative x axis

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