Question

Find an equation of the tangent line to the curve at the given point y=ln(x^2-4x 1),(4,0).

Answer 1

The equation of tangent line to the curve y = ln(x^2 - 4x + 1) is y = 4x -16.

According to the given question.

We have a curve y = ln(x^2 - 4x + 1)

Let the equation of tangent line to the curve y = ln(x^2 - 4x + 1) be

y = mx + b

Where, m is the slope of the curve.

So, the slope of the curve m, y = ln(x^2 - 4x + 1) at x = 4 is given by

m = dy/dx

⇒ m = d(ln(x^2 - 4x + 1))/dx

$\implies m = \frac{1}{(x^{2} -4x+1)}\frac{d}{dx}(x^{2} -4x+1)$

$\implies m = \frac{1}{(x^{2} -4x+1)} (2x -4)$

$\implies m = \frac{1}{(16-16+1)} (8-4)$       (at x = 4)

⇒ m = 1(4)

⇒ m = 4

Substitute the value of m in y = mx + b

⇒ y = 4x + b

Since, the above tangent line is passing through (4, 0). So, the point (4, 0) must satisfy y = 4x + b.

⇒ 0 = 4(4) + b

⇒ b = -16

Therefore, the equation of tangent line to the curve y = ln(x^2 - 4x + 1) is given by

y = mx + b

⇒ y = 4x -16        (by substituting the vale of m and b in y = mx+b)

Hence, the equation of tangent line to the curve y = ln(x^2 - 4x + 1) is y = 4x -16.

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