Question -
Answer -
(i) y2 = 8x
Given:
Parabola = y2 = 8x
Now by comparing with the actual parabola y2 =4ax
Then,
4a = 8
a = 8/4 = 2
So, the vertex is (0, 0)
The focus is (a, 0) = (2, 0)
The equation of the axis is y = 0.
The equation of the directrix is x = – a i.e., x = – 2
The length of the latus rectum is 4a = 8.
(ii) 4x2 + y = 0
Given:
Parabola => 4x2 + y = 0
Now by comparing with the actual parabola y2 =4ax
Then,
4a = ¼
a = 1/(4 × 4)
= 1/16
So, the vertex is (0, 0)
The focus is = (0, -1/16)
The equation of the axis is x = 0.
The equation of the directrix is y = 1/16
The length of the latus rectum is 4a = ¼
(iii) y2 – 4y – 3x + 1 = 0
Given:
Parabola y2 – 4y – 3x + 1 = 0
y2 – 4y = 3x – 1
y2 – 4y + 4 = 3x + 3
(y – 2)2 = 3(x + 1)
Now by comparing with the actual parabola y2 =4ax
Then,
4b = 3
b = ¾
So, the vertex is (-1, 2)
The focus is = (3/4 – 1, 2) = (-1/4, 2)
The equation of the axis is y – 2 = 0.
The equation of the directrix is (x – c) = -b
(x – (-1)) = -3/4
x = -1 – ¾
= -7/4
The length of the latus rectum is 4b = 3
(iv) y2 – 4y + 4x = 0
Given:
Parabola y2 – 4y + 4x = 0
y2 – 4y = – 4x
y2 – 4y + 4 = – 4x + 4
(y – 2)2 = – 4(x – 1)
Now by comparing with the actual parabola y2 =4ax => (y – a)2 = – 4b(x – c)
Then,
4b = 4
b = 1
So, the vertex is (c, a) = (1, 2)
The focus is (b + c, a) = (1-1, 2) = (0, 2)
The equation of the axis is y – a = 0 i.e., y – 2 = 0
The equation of the directrix is x – c = b
x – 1 = 1
x = 1 + 1
= 2
Length of latus rectum is 4b = 4
(v) y2 + 4x + 4y – 3 = 0
Given:
The parabola y2 + 4x + 4y – 3 = 0
y2 + 4y = – 4x + 3
y2 + 4y + 4 = – 4x + 7
(y + 2)2 = – 4(x – 7/4)
Now by comparing with the actual parabola y2 =4ax => (y – a)2 = – 4b(x – c)
Then,
4b = 4
b = 4/4 = 1
So, The vertex is (c, a) = (7/4, -2)
The focus is (- b + c, a) = (-1 + 7/4, -2) = (3/4, -2)
The equation of the axis is y – a = 0 i.e., y + 2 = 0
The equation of the directrix is x – c = b
x – 7/4 = 1
x = 1 + 7/4
= 11/4
Length of latus rectum is 4b = 4.