Question

A parabola having equation 4y²+12x-20y+67=0 has its origin shifted to a point (h,k) such that the equation becomes the standard equation y²=4aX.

Find h,k

Find vertex latus rectum, focus, axis , equation of directrix in standard form and transform them with respect to original axes.

Answer 1

Equation of the parabola is given by 4y² + 12x - 20y + 67 = 0
Dividing throughout by 4 to make the coefficient of y² unity, we get
y² + 3x - 5y + 67/4 = 0
y² -5y = -3x-67/4
So completing the whole square form add 25/4 to both side
So y² -5y +25/4 = -3x -67/4 +25/4
Or (y-5/2 )²= -3x -42/4
Or (y-5/2 )² = -3(x+7/2)
So the origin is shifted to 5/2 and -7/2, such that the new coordinates will be (h,k)
Hence h = (x+7/2) and k = (y-5/2)
k² = 3h

So this is of the form y² = 4ax, hence 4a = 3, or a = 3/4

Vertex:
According to new axis the coordinate will be (0,0)
So with respect to old axis (x+7/2) = 0, or x = -7/2 and 0 = (y-5/2), y = 5/2
Hence the vertex is (-7/2 , 5/2)

Focus:
With respect to new axis the focus is (-a ,0)
So h = (x+7/2), so -3/4 = x+7/2
Hence x = -17/4
k = (y-5/2), so y = 5/2
So focus is ( -17/4 , 5/2 )

Axis:
With respect to the new axes
k = 0
With respect to old axes
y = k + 5/2
= 0 + 5/2
y = 5/2
Or 2y - 5 = 0

Directrix: With respect to the new axes h = a = 3/4 With respect to the old axes x = h - 7/2 = 3/4 - 7/2 = -11/4