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प्रश्न
Find the equations of the tangents to the curve y = `- (x + 1)/(x - 1)` which are parallel to the line x + 2y = 6
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उत्तर
Curse is y = `(x + 1)/(x - 1)`
DIfferentiating w.r.t. 'x'
`("d"y)/("d"x) = ((x - 1)(1) - (x + 1)(1))/(x - 1)^2`
Slope of the tangent 'm'
= `(x - 1 - x - 1)/(x - 1)^2`
= `- 2/(x - 1)^2`
Given line is x + 2y = 6
Slope of the line = ` 1/2`
Since the tangent is parallel to the line, then the slope of the tangent is `- 1/2`
∴ `("d"y)/("d"x) = 2/(x 1)^2 = - 1/2`
(x – 1)2 = 4
x – 1 = ± 2
x = – 1, 3
When x = – 1, y = 0
⇒ point is (– 1, 0)
When x = 3, y = 2
⇒ point is (3, 2)
Equation of tangent with slope `- 1/2` and at the point (– 1, 0) is
y – 0 = `- 1/2(x + 1)`
2y = – x – 1
⇒ x + 2y + 1 = 0
Equation of tangent with slope ` 1/2` and at the point (3, 2) is 2
y – 2 = `- 1/2 (x - 3)`
2y – 4 = – x + 3
x + 2y – 7 = 0.
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