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