Question

Find the vertex, axis, focus and latus rectum for following parabola:

(i) y² = 8x + 8y

(ii) x² + 2y = 8x – 7

Answer 1

(i) y² = 8x + 8y

⇒ y² – 8y = 8x

⇒ y² – 2 × 4 × y + 4² = 8x + 4²

⇒ (y – 4)² = 8x + 16

⇒ (y – 4)² = 8(x + 2) …(i)

For replacing origin at point (4, – 2), put x + 2 and y – 4 = Y

Y² = 8W

⇒ Y² = 4.2.X …(i)

Which is of the form of parabola y² = 4ax where a = 2 and axis X = 0

Coordinates of vertex = (0, 0), coordinates of focus = (0,-1)

Length of Latus Rectum = 4 × 2 = 8

for given parabola (i), put value of X and Y in results.

X = 0 ⇒ x + 2 – 0 ⇒ x = -2

Y = 0 ⇒ y – 4 = 0 ⇒ y = 4

Thus coordinates of vertex = (- 2, 4)

Coordinates of focus

X = 0 ⇒ x + 2 = 2, x = 0

Y = 0 ⇒ y – 4 = 0, y = 4

Thus, coordinates of focus = (0, 4)

Axis Y = 0 ⇒ y – 4 = 0 ⇒ y = 4

Latus rectum = 4a = 4 × 2 = 8

(ii) x² + 2y = 8x – 7

x² – 8x = – 2y – 7

⇒ x² – 2 × 4 × x + 42 = -2y – 7 + 4²

⇒ (x – 4)² = – 2y – 7 + 16

⇒ (x – 4)² = – 2y + 9

In parabola X² = 4ay

Axis X = 0, then x – 4 = 0 

⇒ x = 4

Vertex (0, 0), then

x – 4 = 0 

⇒ x = 4