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# Find the equations of all lines having slope 0 which are tangent to the curve y = 1/(x^{2} - 2x + 3)

**Solution:**

The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line.

For a curve y = f(x) containing the point (x_{1},y_{1}) the equation of the tangent line to the curve at (x_{1},y_{1}) is given by

y − y_{1} = f′(x_{1}) (x − x_{1})

The given curve is

y = 1/(x^{2} - 2x + 3)

The slope of the tangents to the given curve is given by,

dy/dx

= d/dx (1/(x^{2} - 2x + 3))

= - (2x - 2)/(x^{2} - 2x + 3)^{2}

= - 2(x - 1)/(x^{2} - 2x + 3)^{2}

Given that the slope of the tangent is 0.

So, we have:

- 2(x - 1)/(x^{2} - 2x + 3)^{2} = 0

⇒ - 2(x - 1) = 0

⇒ x = 1

When,

Then,

y = 1/(1 - 2 + 3)

= 1/2

Hence, the equation of the tangent through (1, 1/2) is given by,

y - 1/2 = 0 (x - 1)

⇒ y - 1/2 = 0

⇒ y = 1/2

Thus,

the equation of the line is y - 1/2 = 0

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 12

## Find the equations of all lines having slope 0 which are tangent to the curve y = 1/(x^{2} - 2x + 3)

**Summary:**

The equation of all lines having slope 0 which are tangent to the curve y = 1/(x^{2} - 2x + 3) is y - 1/2 = 0.The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line

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