Find the standard form of the equation of the parabola with the given characteristics. Vertex: (5, 1); Focus (2, 1)

Solution:

The parabola with the Vertex(5, 1) and Focus F(2.1)  is presented below:

Please Note: The horizontal axis is y and the vertical axis is x

For any point P(x, y) to lie on the parabola the distance between the focus F(2,1) and P(x, y) is equal to the distance between P(x, y) and Q(x, y) which lies on the directrix. Using this input we get

Distance PF  = Distance PQ

√(x - 2)² + (y -1)² =  √(x - 8)² + (y - y)² 

Squaring both the sides we have,

(x - 2)² + (y - 1)² = (x - 8)²

 (y - 1)² =  (x - 8)² - (x - 2)²

Using the formula a² - b² =  (a - b)(a + b) on the RHS of the above equation we have,

(y - 1)²  =  (x - 8 - x + 2)(x - 8 + x - 2)

(y - 1)² = (-6)(2x - 10)

(y -1)² = -12x + 60

The final equation for the standard form of parabola is written as

12x = 60 - (y -1)²

x =  (60 - (y -1)² ) / 12

x =  5  -  ((y -1)² / 12)

The above equation of the parabola can be verified by substituting the y-coordinate of the vertex (5,1) in the equation.

Find the standard form of the equation of the parabola with the given characteristics. Vertex: (5, 1); Focus (2, 1)

Summary:

The standard form of the equation of the parabola with the given characteristics. Vertex: (5, 1); Focus (2, 1) is x =  5  -  ((y - 1)² / 12)