Find the equation of the parabola with a focus (2, 5) and directrix of y = 3.A.)y=x^2/4 + x + 5B.)y=x^2/4 + x -5C.)y=x^2/4 -x - 5D.)y=x^2/4 - x + 5
Accepted Solution
A:
Answer:
One form of an equation is given by y=a(x−h
)
2
+k
y=a(x−h)2+k
where (h,k)
(h,k)
the coordinates of the vertex and (h,k+
1
4a
)
The parabola is symmetric with respect to y=x
y=x
and can be viewed as a standard downward parabola with a rotation of 45 degrees clockwise. So, follow the steps below to obtain its equation.
1) The length between the focus and vertex is f=
4
2
+
4
2
−
−
−
−
−
−
√
=4
2
–
√
f=42+42=42
. The standard equation is y=−
1
4f
x
2
y=−14fx2
.
2) Shift the vertex to (-2, -2),
y+2=−
1
16
2
–
√
(x+2
)
2
y+2=−1162(x+2)2
3) Rotate the equation -45-degrees x→
1
2
√
(x−y)
x→12(x−y)
, y→−
1
2
√
(x+y)
y→−12(x+y)
to get
−
1
2
–
√
(x+y)+2=−
1
16
2
–
√
(
1
2
–
√
(x−y)+2)
2
Step-by-step explanation: