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प्रश्न
Find the equation of tangent and normal to the following curve.
y = x3 - x2 - 1 at the point whose abscissa is -2.
उत्तर
Equation of the curve is y = x3 - x2 - 1 ...(i)
Differentiating w.r.t. x, we get
`"dy"/"dx" = 3"x"^2 - 2"x"`
If x = - 2, ....[Given]
Putting the value of x in (i), we get
y = (- 2)3 - (- 2)2 - 1 = - 8 - 4 - 1 = - 13
∴ Point is P (x1, y1) ≡ (-2, -13)
Slope of tangent at (- 2,- 13) is
`("dy"/"dx")_((-2, -13)` = 2(-2)2 - 2(-2) = 12 + 4 = 16
Equation of tangent at (- 2, -13) is
y - y1 = `("dy"/"dx")_(("x" = -2)` (x - x1)
∴ y - (- 13) = 16 [x - (- 2)]
∴ y + 13 = 16x + 32
∴ 16x - y + 19 = 0
Slope of normal at (- 2, - 13) is `(-1)/("dy"/"dx")_((-2, -13)` = `- 1/16`
Equation of normal at (-2, - 13) is
∴ y + 13 = `(-1)/16`(x + 2)
∴ x + 16y + 210 = 0
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