1)Find the focus and directrix of the parabola y^2= -32x?

Question

2)Find the equation of the parabola with focus (2, 3) and directrix y = -1

Answer 1

1)

 

The equation of the parabola : y² = -32x

The equation is in the form of  Y² = 4aX

where : 

 X = x , Y = y, 4a = -32,------a = -8

 Focus

( X = a, Y = 0 ) ≡ ( x = (-8), y = 0 ) ≡ ( -8, 0 )

 Directrix : X = -a ∴ x = -(-8) ∴ x = 8 

 

Answer 2

2)

 Parabola is the locus of points that are equidistant to the focus and the directrix.
If P(x,y) is a point on the parabola, then 
 The square of its distance to the focus F(2,3) is:
PF² = (x- 2)² + (y- 3)²
The square of its distance PH to the directrix y = -1 is:
PH² = (y + 1)²

Since PF = PH
(x - 2)² + (y-3)² = (y + 1)²
(x- 2)² = (y + 1)²- (y - 3)²
x² - 4x + 4 = (y + 1 + y - 3)(y + 1 - y + 3)
x² - 4x + 4 =8y-8 
8y = x² - 4x + 12