Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1. f(x) = −one twelfth(x − 5)2 + 2
f(x) = one twelfth(x − 5)2 + 2
f(x) = −one twelfth(x + 5)2 + 2
f(x) = one twelfth(x + 5)2 + 2

The equation of the parabola with a focus at (-5,5) and a directrix of y = -1 is .

Step-by-step explanation:

From statement we understand that parabola has its axis of symmetry in an axis parallel to y-axis. According to Analytical Geometry, the minimum distance between focus and directrix equals to twice the distance between vertex and any of endpoints.

If endpoints are (-5, 5) and (-5, -1), respectively, then such distance (), dimensionless, is calculated by means of the Pythagorean Theorem:

And the location of the vertex (), dimensionless, which is below the focus, is:

(1)

Where:

- Focus, dimensionless.

- Vector distance, dimensionless.

If we know that and , then the location of the vertex is:

In addition, we define a parabola by the following expression:

(2)

Where:

, - Coordinates of the vertex, dimensionless.

- Distance of the focus with respect to vertex, dimensionless.

If we know that , and , then the equation of the parabola is:

The equation of the parabola with a focus at (-5,5) and a directrix of y = -1 is .

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this