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Here 4a = 16 or a = 4.
The focus of the parabola lies on the y-axis and its
opening is downward. Coordinates of the focus = (0, -4).
Equation of its axis is x = 0
Length of the latusrectum is 16 and y = 0 is tangent to
the parabola at its vertex. The shape of the curve is as shown
in the figure.

Example 2. Find an equation of the parabola whose focus is F (-3, 4) and directrix is
3x - 4y + 5 = 0.

Solution: Let P(x , y) be a point on the parabola. Lentgh of the perpendicular PM from
P(x , y) to the directrix 3x - 4y + 5 = 0 is

PM 3x - 4y + 5
32 + (-4)2

=By definition, PF P=M or PF 2 PM 2

or (x + 3)2 + ( y - 4)2 =(3x - 4 y + 5)2
25

or 25(x2 + 6x + 9 + y2 - 8y + 16) = 9x2 + 16y2 + 25 - 24xy + 30x - 40y

or 16x2 +24xy + 9y2 + 120x - 160y +600 = 0

is an equation of the required parabola.

Example 3. Analyze the parabola
x2 - 4x - 3y + 13 = 0
and sketch its graph.

Solution. The given equation may be written as version: 1.1
x2 - 4x + 4 = 3y - 9 (1)
or (x - 2)2 = 3(y - 3)
Let x - 2 = X , y - 3 = Y (2)
The equation (2) becomes X2 = 3Y (3)

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