Find the standard form of the equation of the parabola with the given characteristics? Vertex: (-2, 1) Directrix: x = 1

Solution:

If the vertex of the parabola is (-2, 10) and the directrix is x = 1 then the Focus of the parabola is at the point F(-5, 1) because the distance between the vertex and focus is the same as the distance between the vertex and the directrix.

To find the standard form of the equation of the parabola we identify a point P(x, y) on the parabola. The point P(x, y) will lie on the parabola only and only if its distance from the point F(-5, 1) is the same as its distance from the point (1, y) on the directrix.

Therefore we can write:

Distance between F(-5, 1) and P(x, y)  = Distance between P(x, y) end point  (1, y) on the directrix

\(\sqrt{(x+5)^{2} + (y-1)^{2}} = \sqrt{(x-1)^{2}}\)

Squaring both sides we have 

\({(x+5)^{2} + (y-1)^{2}} = (x-1)^{2}\)

\((y-1)^{2} = (x-1)^{2} - (x+5)^{2}\)

We know that a² - b²  = (a + b)(a - b), therefore 

(y - 1)² = ( x -1 + x + 5)(x - 1 - x -5)

(y - 1)² = (2x + 4)(-6)

(y - 1)² = -24 - 12x

-12x  =   24 + (y - 1)²

x = -2  -  (y - 1)² / 12

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Find the standard form of the equation of the parabola with the given characteristics? Vertex: (-2, 1) Directrix: x = 1

Summary:

The standard form of the equation of the parabola with the given characteristics i.e.  Vertex: (-2, 1) Directrix: x = 1 is given by x = -2  -  (y - 1)²  / 12