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Question
Find the equations of the tangents to the curve y = 1 + x3 for which the tangent is orthogonal with the line x + 12y = 12
Solution
Curve y = 1 + x3
Differentiating w.r.t ‘x’
Slope of the tangent ‘m’ = `("d"y)/("d"x)` = 3x2
Given line is x + 12y = 12
Slope of the line is `- 1/12`
Since the tangent is orthogonal with the line, the slope of the tangent is 12.
∴ `("d"y)/("d"x)` = 12
i.e 3x2 = 12
x2 = 4
x = ± 2
When x = 2
y = 1 + 8
= 9
⇒ point is (2, 9)
When x = – 2
y = 1 – 8
= – 7
⇒ point is (– 2, – 7)
Equation of tangent with slope 12 and at the j point (2, 9) is
y – 9 = 12(x – 2)
y – 9 = 12x – 24
12x – y – 15 = 0
Equation of tangent with slope 12 and at the point (– 2, – 7) is
y + 7 = 12(x + 2)
y + 7 = 12x + 24
12x – y + 17 = 0
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