If the tangent to the curve, y = x3 + ax – b at the point (1, –5) is perpendicular to the line, – x + y + 4 = 0, then which one of the following points lies on the curve?

Engineering

Mathematics

Tangent and Normal

Question

If the tangent to the curve, y = x³ + ax – b at the point (1, –5) is perpendicular to the line, – x + y + 4 = 0, then which one of the following points lies on the curve?

(–2, 2)

(2, –2)

(–2, 1)

(2, –1)

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y = x³ + ax – b

(1, – 5) lies on the curve

⇒ –5 = 1 + a – b ⇒ a – b = –6 …….(i)

Also, y' = 3x² + a

y '(1,– 5) = 3 + a (slope of tangent)

$∵$ this tangent is $⊥$ to –x + y + 4 = 0

⇒ (3 + a)(1) = –1

⇒ a = –4 ………(ii)

By (i) and (ii) : a = –4, b = 2

$∴$ y = x³ – 4x – 2, (2, –2) lies on this curve.