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Find the equations of all lines having slope 0 which are tangent to the curve y = 1/(x2 - 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 (x1,y1) the equation of the tangent line to the curve at (x1,y1) is given by
y − y1 = f′(x1) (x − x1)
The given curve is
y = 1/(x2 - 2x + 3)
The slope of the tangents to the given curve is given by,
dy/dx
= d/dx (1/(x2 - 2x + 3))
= - (2x - 2)/(x2 - 2x + 3)2
= - 2(x - 1)/(x2 - 2x + 3)2
Given that the slope of the tangent is 0.
So, we have:
- 2(x - 1)/(x2 - 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/(x2 - 2x + 3)
Summary:
The equation of all lines having slope 0 which are tangent to the curve y = 1/(x2 - 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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