i)

y² = k/(x - 1) , k is a constant.

When y = 1/7 and x = 8,

(1/7)² = k/(8 - 1)
1/49 = k/7

k = 1/7

ii)

So, y² = 1 / 7(x - 1)

Given:

when x = a, y = b

b² = 1 / 7(a - 1)


When y is doubled, the new y = 2b

new y² = (2b)² = 4b² = 4 / 7(a - 1)

So,

4 / 7(a - 1) = 1 / 7(x - 1)

4 / (a - 1) = 1 / (x - 1)

Invert the fractions,

(a - 1) / 4 = x - 1

x = (a - 1) / 4 + 1

x = ¼a + ¾

Alternatively,

y² = 1 / 7(x - 1)

x - 1 = 1 / 7y²

x = 1 + 1 / 7y²

When x = a, y = b,

a = 1 + 1 / 7b²

When y is doubled, it becomes 2b.

x = 1 + 1 / 7(2b²)

x = 1 + 1 / 7(4b²)

x = 1 + ¼ (1/7b²)

x = ¾ + ¼ + ¼ (1/7b²)

x = ¾ + ¼(1 + 1/7b²)

x = ¾ + ¼a

Another alternative (based on logic)

When y is doubled, it is twice its original value.

So y² would be 4 times its original value.

Since y² is inversely proportional to (x - 1),
then (x - 1) would be ¼ of its original value.

Since the original x = a,
then the original x - 1 = a - 1

So,

The new x - 1 = ¼ (a - 1)

new x = ¼a - ¼ + 1
new x = ¼a + ¾