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Find the equation of tangent and normal to the curve y = x3 + 4x2 at (–1, 3). |
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Answer» Given curve equation is y = x3 + 4x2 ——— (1) dy/dx = 3x2 + 8x m = (dy/dx)(–1, 3) = 3(–1)2 + 8(–1) = 3 – 8 = –5 The equation of the tangent to the curve (1) at (–1, 3) is y – y1 = m(x – x1) y – 3 = –5(x + 1) y – 3 = –5x – 5 ⇒ 5x + y + 2 = 0 The equation of the normal to the curve (1) at (–1, 3) is y – y1 = −1/m(x – x1) ⇒ y – 3 = − 1/5(x + 1) ⇒ 5y – 15 = x + 1 ⇒ x – 5y + 16 = 0 |
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