Question

Find the points on the curve y² – 4xy = x² + 5 for which the tangent is horizontal

Answer 1

y² – 4xy = x² + 5  .........(1)

Differentiating w.r.t. ‘x’

`2y  ("d"y)/("d"x) - 4 (x  ("d"y)/("d"x) + y.1)` = 2x

`2y  ("d"y)/("d"x) - 4x ("d"y)/("d"x) - 4y` = 2x

`("d"y)/("d"x) (2y - 4x)` = 2x + 4y

∴ `("d"y)/("d"x) = (2(x + 2y))/(2(y - 2x))`

= `(x + 2y)/(y - 2x)`

When the tangent is horizontal(Parallel to X-axis) then slope of the tangent is zero.

`("d"y)/("d"x)` = 0

⇒ `(x + 2y)/(y - 2x)` = 0

⇒ x + 2y = 0

x = – 2y

Substituting in (1)

y² – 4 (– 2y) y = (– 2y)² + 5

y² + 8y² = 4y² + 5

5y² = 5

⇒ y² = 1

y = ±1

When y = 1, x = – 2

When y = – 1, x = 2

∴ The points on the curve are (– 2, 1) and (2, –1).