jmc

Question

The graph of a parabola has the following properties:

∙

It passes through the point

(1, 5) .

∙

The

y

-coordinate of the focus is 3.

∙

Its axis of symmetry is parallel to the

x

-axis.

∙

Its vertex lies on the

y

-axis.

Express the equation of the parabola in the form

a x^{2} + b x y + c y^{2} + d x + ey + f = 0,

where

a,

b,

c,

d,

e,

f

are integers,

c

is a positive integer, and

g cd (∣ a ∣, ∣ b ∣, ∣ c ∣, ∣ d ∣, ∣ e ∣, ∣ f ∣) = 1.

Accepted Answer

Since the axis of symmetry is parallel to the

x

-axis, and the

y

-coordinate of the focus is 3, the

y

-coordinate of the vertex is also 3. Since the vertex lies on the

y

-axis, it must be at

(0, 3) .

Hence, the equation of the parabola is of the form

x = k (y - 3)^{2} .

Since the graph passes through

(1, 5),

we can plug in

x = 1

and

y = 5,

to get

1 = 4 k,

so

k = \frac{1}{4} .

Hence, the equation of the parabola is

x = \frac{1}{4} (y - 3)^{2},

which we write as

y^{2} - 4 x - 6 y + 9 = 0 .