Given:

The curve is = 1

Differentiating the above w.r.t x, we get the The Slope of a tangent,

⇒ = 0

Cross multiplying we get,

⇒ = 0

⇒ 50x + 8y = 0

⇒ 8y = – 50x

⇒ =

⇒ = ...(1)

(i)

Since, the tangent is parallel to x – axis

⇒ = tan(0) = 0 ...(2)

tan(0) = 0

= The Slope of the tangent = tan

From (1) & (2),we get,

⇒ = 0

⇒ – 25x = 0

⇒ x = 0

Substituting x = 0 in = 1,

= 1

⇒ y² = 25

⇒ y = ±5

Thus, the required point is (0,5) & (0, – 5)